Proof By Induction Trigonometry Questions, Answers and Solutions
a - answer s - solution v - video
Question 1
Prove by mathematical induction for $n\ge1$ and $n\in\mathbb{Z}^+$:
a) $\cos x\times\cos 2x\times\dots\times\cos(2^{n-1}x)=\frac{\sin(2^nx)}{2^n\sin(x)}$
a
Answer and Solution are the same for proofs
To prove: $\cos x\times\cos 2x\times\dots\times\cos(2^{n-1}x)=\frac{\sin(2^nx)}{2^n\sin(x)}$ for $n\ge1$ and $n\in\mathbb{Z}^+$:
Step 1 - Prove true for $n=1$
$LHS=\cos x$
$RHS=\frac{\sin(2x)}{2\sin x}$
$=\frac{2\sin x\cos x}{2\sin x}$
$=\cos x$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\cos x\times\cos 2x\times\dots\times\cos(2^{k-1}x)=\frac{\sin(2^kx)}{2^k\sin(x)}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\cos x\times\cos 2x\times\dots\times\cos(2^{(k+1)-1}x)=\frac{\sin(2^{k+1}x)}{2^{k+1}\sin(x)}$
$LHS=\cos x\times\cos 2x\times\dots\times\cos(2^{k-1}x)\times\cos(2^kx)$
$=\frac{\sin(2^kx)}{2^k\sin(x)}\times\cos(2^kx)$ (using Step 2)
$=\frac{\sin(2^kx)\cos(2^kx)}{2^k\sin(x)}$
$=\frac{2\sin(2^kx)\cos(2^kx)}{2^{k+1}\sin(x)}$
$=\frac{\sin(2\times2^kx)}{2^{k+1}\sin(x)}$
$=\frac{\sin(2^{k+1}x)}{2^{k+1}\sin(x)}$
$=RHS$
Step 4 - Conclusion
If true for $n=k$, then we have proven it is true for $n=k+1$.
Also, as true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
s
Answer and Solution are the same for proofs
To prove: $\cos x\times\cos 2x\times\dots\times\cos(2^{n-1}x)=\frac{\sin(2^nx)}{2^n\sin(x)}$ for $n\ge1$ and $n\in\mathbb{Z}^+$:
Step 1 - Prove true for $n=1$
$LHS=\cos x$
$RHS=\frac{\sin(2x)}{2\sin x}$
$=\frac{2\sin x\cos x}{2\sin x}$
$=\cos x$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\cos x\times\cos 2x\times\dots\times\cos(2^{k-1}x)=\frac{\sin(2^kx)}{2^k\sin(x)}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\cos x\times\cos 2x\times\dots\times\cos(2^{(k+1)-1}x)=\frac{\sin(2^{k+1}x)}{2^{k+1}\sin(x)}$
$LHS=\cos x\times\cos 2x\times\dots\times\cos(2^{k-1}x)\times\cos(2^kx)$
$=\frac{\sin(2^kx)}{2^k\sin(x)}\times\cos(2^kx)$ (using Step 2)
$=\frac{\sin(2^kx)\cos(2^kx)}{2^k\sin(x)}$
$=\frac{2\sin(2^kx)\cos(2^kx)}{2^{k+1}\sin(x)}$
$=\frac{\sin(2\times2^kx)}{2^{k+1}\sin(x)}$
$=\frac{\sin(2^{k+1}x)}{2^{k+1}\sin(x)}$
$=RHS$
Step 4 - Conclusion
If true for $n=k$, then we have proven it is true for $n=k+1$.
Also, as true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
v
To prove: $\cos x\times\cos 2x\times\dots\times\cos(2^{n-1}x)=\frac{\sin(2^nx)}{2^n\sin(x)}$ for $n\ge1$ and $n\in\mathbb{Z}^+$:
Step 1 - Prove true for $n=1$
$LHS=\cos x$
$RHS=\frac{\sin(2x)}{2\sin x}$
$=\frac{2\sin x\cos x}{2\sin x}$
$=\cos x$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\cos x\times\cos 2x\times\dots\times\cos(2^{k-1}x)=\frac{\sin(2^kx)}{2^k\sin(x)}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\cos x\times\cos 2x\times\dots\times\cos(2^{(k+1)-1}x)=\frac{\sin(2^{k+1}x)}{2^{k+1}\sin(x)}$
$LHS=\cos x\times\cos 2x\times\dots\times\cos(2^{k-1}x)\times\cos(2^kx)$
$=\frac{\sin(2^kx)}{2^k\sin(x)}\times\cos(2^kx)$ (using Step 2)
$=\frac{\sin(2^kx)\cos(2^kx)}{2^k\sin(x)}$
$=\frac{2\sin(2^kx)\cos(2^kx)}{2^{k+1}\sin(x)}$
$=\frac{\sin(2\times2^kx)}{2^{k+1}\sin(x)}$
$=\frac{\sin(2^{k+1}x)}{2^{k+1}\sin(x)}$
$=RHS$
Step 4 - Conclusion
If true for $n=k$, then we have proven it is true for $n=k+1$.
Also, as true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
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Answer by: ada
To prove: $\cos x\times\cos 2x\times\dots\times\cos(2^{n-1}x)=\frac{\sin(2^nx)}{2^n\sin(x)}$ for $n\ge1$ and $n\in\mathbb{Z}^+$:
Step 1 - Prove true for $n=1$
$LHS=\cos x$
$RHS=\frac{\sin(2x)}{2\sin x}$
$=\frac{2\sin x\cos x}{2\sin x}$
$=\cos x$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\cos x\times\cos 2x\times\dots\times\cos(2^{k-1}x)=\frac{\sin(2^kx)}{2^k\sin(x)}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\cos x\times\cos 2x\times\dots\times\cos(2^{(k+1)-1}x)=\frac{\sin(2^{k+1}x)}{2^{k+1}\sin(x)}$
$LHS=\cos x\times\cos 2x\times\dots\times\cos(2^{k-1}x)\times\cos(2^kx)$
$=\frac{\sin(2^kx)}{2^k\sin(x)}\times\cos(2^kx)$ (using Step 2)
$=\frac{\sin(2^kx)\cos(2^kx)}{2^k\sin(x)}$
$=\frac{2\sin(2^kx)\cos(2^kx)}{2^{k+1}\sin(x)}$
$=\frac{\sin(2\times2^kx)}{2^{k+1}\sin(x)}$
$=\frac{\sin(2^{k+1}x)}{2^{k+1}\sin(x)}$
$=RHS$
Step 4 - Conclusion
If true for $n=k$, then we have proven it is true for $n=k+1$.
Also, as true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
b) $\cos\left(\frac{x}{2}\right) + \cos\left(\frac{3x}{2}\right)+\cdots + \cos\left(\frac{(2n-1)x}{2}\right) = \frac{\sin nx}{2\sin\left(\frac x 2\right)}$
a
Answer and Solution are the same for proofs
To prove: $\cos\left(\frac{x}{2}\right) + \cos\left(\frac{3x}{2}\right)+\cdots + \cos\left(\frac{(2n-1)x}{2}\right) = \frac{\sin nx}{2\sin\left(\frac x 2\right)}$ for $n\ge1$ and $n\in\mathbb{Z}^+$:
Step 1 - Prove true for $n=1$
$LHS=\cos\left(\frac x2\right)$
$RHS=\frac{\sin x}{2\sin\left(\frac x 2\right)}$
$=\frac{2\sin\left(\frac x2\right)\cos\left(\frac x2\right)}{2\sin\left(\frac x 2\right)}$
$=\cos\left(\frac x2\right)$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\cos\left(\frac{x}{2}\right) + \cos\left(\frac{3x}{2}\right)+\cdots + \cos\left(\frac{(2k-1)x}{2}\right) = \frac{\sin kx}{2\sin\left(\frac x 2\right)}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\cos\left(\frac{x}{2}\right) + \cos\left(\frac{3x}{2}\right)+\cdots + \cos\left(\frac{(2k-1)x}{2}\right) + \cos\left(\frac{(2(k+1)-1)x}{2}\right) = \frac{\sin(k+1)x}{2\sin\left(\frac x 2\right)}$
$LHS=\cos\left(\frac{x}{2}\right) + \cos\left(\frac{3x}{2}\right)+\cdots + \cos\left(\frac{(2k-1)x}{2}\right) + \cos\left(\frac{(2k+1)x}{2}\right)$
$=\frac{\sin kx}{2\sin\left(\frac x 2\right)}+ \cos\left(\frac{(2k+1)x}{2}\right)$ (using Step 2)
$=\frac{\sin kx + 2\sin\left(\frac x 2\right)\cos\left(kx+\frac x 2\right)}{2\sin\left(\frac x 2\right)}$
$=\frac{\sin kx + 2\sin\left(\frac x 2\right)\left[\cos kx\cos\left(\frac x 2\right)-\sin kx\sin\left(\frac x 2\right)\right]}{2\sin\left(\frac x 2\right)}$
$\frac{\sin kx \left[1-2\sin^2\left(\frac x 2\right)\right]+2\sin\left(\frac x 2\right)\cos\left(\frac x 2\right)\cos kx}{2\sin\left(\frac x 2\right)}$
$=\frac{\sin kx\cos x+\sin x\cos kx}{2\sin\left(\frac x 2\right)}$
$=\frac{\sin(kx+x)}{2\sin\left(\frac x 2\right)}$
$=\frac{\sin(k+1)x}{2\sin\left(\frac x 2\right)}$
$=RHS$
Step 4 - Conclusion
If true for $n=k$, then we have proven it is true for $n=k+1$.
Also, as true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
s
Answer and Solution are the same for proofs
To prove: $\cos\left(\frac{x}{2}\right) + \cos\left(\frac{3x}{2}\right)+\cdots + \cos\left(\frac{(2n-1)x}{2}\right) = \frac{\sin nx}{2\sin\left(\frac x 2\right)}$ for $n\ge1$ and $n\in\mathbb{Z}^+$:
Step 1 - Prove true for $n=1$
$LHS=\cos\left(\frac x2\right)$
$RHS=\frac{\sin x}{2\sin\left(\frac x 2\right)}$
$=\frac{2\sin\left(\frac x2\right)\cos\left(\frac x2\right)}{2\sin\left(\frac x 2\right)}$
$=\cos\left(\frac x2\right)$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\cos\left(\frac{x}{2}\right) + \cos\left(\frac{3x}{2}\right)+\cdots + \cos\left(\frac{(2k-1)x}{2}\right) = \frac{\sin kx}{2\sin\left(\frac x 2\right)}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\cos\left(\frac{x}{2}\right) + \cos\left(\frac{3x}{2}\right)+\cdots + \cos\left(\frac{(2k-1)x}{2}\right) + \cos\left(\frac{(2(k+1)-1)x}{2}\right) = \frac{\sin(k+1)x}{2\sin\left(\frac x 2\right)}$
$LHS=\cos\left(\frac{x}{2}\right) + \cos\left(\frac{3x}{2}\right)+\cdots + \cos\left(\frac{(2k-1)x}{2}\right) + \cos\left(\frac{(2k+1)x}{2}\right)$
$=\frac{\sin kx}{2\sin\left(\frac x 2\right)}+ \cos\left(\frac{(2k+1)x}{2}\right)$ (using Step 2)
$=\frac{\sin kx + 2\sin\left(\frac x 2\right)\cos\left(kx+\frac x 2\right)}{2\sin\left(\frac x 2\right)}$
$=\frac{\sin kx + 2\sin\left(\frac x 2\right)\left[\cos kx\cos\left(\frac x 2\right)-\sin kx\sin\left(\frac x 2\right)\right]}{2\sin\left(\frac x 2\right)}$
$\frac{\sin kx \left[1-2\sin^2\left(\frac x 2\right)\right]+2\sin\left(\frac x 2\right)\cos\left(\frac x 2\right)\cos kx}{2\sin\left(\frac x 2\right)}$
$=\frac{\sin kx\cos x+\sin x\cos kx}{2\sin\left(\frac x 2\right)}$
$=\frac{\sin(kx+x)}{2\sin\left(\frac x 2\right)}$
$=\frac{\sin(k+1)x}{2\sin\left(\frac x 2\right)}$
$=RHS$
Step 4 - Conclusion
If true for $n=k$, then we have proven it is true for $n=k+1$.
Also, as true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
v
To prove: $\cos\left(\frac{x}{2}\right) + \cos\left(\frac{3x}{2}\right)+\cdots + \cos\left(\frac{(2n-1)x}{2}\right) = \frac{\sin nx}{2\sin\left(\frac x 2\right)}$ for $n\ge1$ and $n\in\mathbb{Z}^+$:
Step 1 - Prove true for $n=1$
$LHS=\cos\left(\frac x2\right)$
$RHS=\frac{\sin x}{2\sin\left(\frac x 2\right)}$
$=\frac{2\sin\left(\frac x2\right)\cos\left(\frac x2\right)}{2\sin\left(\frac x 2\right)}$
$=\cos\left(\frac x2\right)$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\cos\left(\frac{x}{2}\right) + \cos\left(\frac{3x}{2}\right)+\cdots + \cos\left(\frac{(2k-1)x}{2}\right) = \frac{\sin kx}{2\sin\left(\frac x 2\right)}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\cos\left(\frac{x}{2}\right) + \cos\left(\frac{3x}{2}\right)+\cdots + \cos\left(\frac{(2k-1)x}{2}\right) + \cos\left(\frac{(2(k+1)-1)x}{2}\right) = \frac{\sin(k+1)x}{2\sin\left(\frac x 2\right)}$
$LHS=\cos\left(\frac{x}{2}\right) + \cos\left(\frac{3x}{2}\right)+\cdots + \cos\left(\frac{(2k-1)x}{2}\right) + \cos\left(\frac{(2k+1)x}{2}\right)$
$=\frac{\sin kx}{2\sin\left(\frac x 2\right)}+ \cos\left(\frac{(2k+1)x}{2}\right)$ (using Step 2)
$=\frac{\sin kx + 2\sin\left(\frac x 2\right)\cos\left(kx+\frac x 2\right)}{2\sin\left(\frac x 2\right)}$
$=\frac{\sin kx + 2\sin\left(\frac x 2\right)\left[\cos kx\cos\left(\frac x 2\right)-\sin kx\sin\left(\frac x 2\right)\right]}{2\sin\left(\frac x 2\right)}$
$\frac{\sin kx \left[1-2\sin^2\left(\frac x 2\right)\right]+2\sin\left(\frac x 2\right)\cos\left(\frac x 2\right)\cos kx}{2\sin\left(\frac x 2\right)}$
$=\frac{\sin kx\cos x+\sin x\cos kx}{2\sin\left(\frac x 2\right)}$
$=\frac{\sin(kx+x)}{2\sin\left(\frac x 2\right)}$
$=\frac{\sin(k+1)x}{2\sin\left(\frac x 2\right)}$
$=RHS$
Step 4 - Conclusion
If true for $n=k$, then we have proven it is true for $n=k+1$.
Also, as true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
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Question by: ada
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To prove: $\cos\left(\frac{x}{2}\right) + \cos\left(\frac{3x}{2}\right)+\cdots + \cos\left(\frac{(2n-1)x}{2}\right) = \frac{\sin nx}{2\sin\left(\frac x 2\right)}$ for $n\ge1$ and $n\in\mathbb{Z}^+$:
Step 1 - Prove true for $n=1$
$LHS=\cos\left(\frac x2\right)$
$RHS=\frac{\sin x}{2\sin\left(\frac x 2\right)}$
$=\frac{2\sin\left(\frac x2\right)\cos\left(\frac x2\right)}{2\sin\left(\frac x 2\right)}$
$=\cos\left(\frac x2\right)$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\cos\left(\frac{x}{2}\right) + \cos\left(\frac{3x}{2}\right)+\cdots + \cos\left(\frac{(2k-1)x}{2}\right) = \frac{\sin kx}{2\sin\left(\frac x 2\right)}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\cos\left(\frac{x}{2}\right) + \cos\left(\frac{3x}{2}\right)+\cdots + \cos\left(\frac{(2k-1)x}{2}\right) + \cos\left(\frac{(2(k+1)-1)x}{2}\right) = \frac{\sin(k+1)x}{2\sin\left(\frac x 2\right)}$
$LHS=\cos\left(\frac{x}{2}\right) + \cos\left(\frac{3x}{2}\right)+\cdots + \cos\left(\frac{(2k-1)x}{2}\right) + \cos\left(\frac{(2k+1)x}{2}\right)$
$=\frac{\sin kx}{2\sin\left(\frac x 2\right)}+ \cos\left(\frac{(2k+1)x}{2}\right)$ (using Step 2)
$=\frac{\sin kx + 2\sin\left(\frac x 2\right)\cos\left(kx+\frac x 2\right)}{2\sin\left(\frac x 2\right)}$
$=\frac{\sin kx + 2\sin\left(\frac x 2\right)\left[\cos kx\cos\left(\frac x 2\right)-\sin kx\sin\left(\frac x 2\right)\right]}{2\sin\left(\frac x 2\right)}$
$\frac{\sin kx \left[1-2\sin^2\left(\frac x 2\right)\right]+2\sin\left(\frac x 2\right)\cos\left(\frac x 2\right)\cos kx}{2\sin\left(\frac x 2\right)}$
$=\frac{\sin kx\cos x+\sin x\cos kx}{2\sin\left(\frac x 2\right)}$
$=\frac{\sin(kx+x)}{2\sin\left(\frac x 2\right)}$
$=\frac{\sin(k+1)x}{2\sin\left(\frac x 2\right)}$
$=RHS$
Step 4 - Conclusion
If true for $n=k$, then we have proven it is true for $n=k+1$.
Also, as true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
c) $\frac12+\cos x+\cos 2x+\dotsb+\cos nx=\frac{\sin\left(n+\frac12\right)x}{2\sin\frac12x}$
a
Answer and Solution are the same for proofs
To prove: $\frac12+\cos x+\cos 2x+\dotsb+\cos nx=\frac{\sin\left(n+\frac12\right)x}{2\sin\frac12x}$
Step 1 - Prove true for $n=1$
$LHS=\frac12+\cos x$
$RHS=\frac{\sin\left(1+\frac12\right)x}{2\sin\frac12x}$
$=\frac{\sin\frac32x}{2\sin\frac12x}$
$=\frac{3\sin\frac12x-4\sin^3\frac12x}{2\sin\frac12x}$
$=\frac{3-4\sin^2\frac12x}{2}$
$=\frac{3-2(1-\cos x)}{2}$
$=\frac{1+2\cos x}{2}$
$=\frac12+\cos x$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\frac12+\cos x+\cos 2x+\dotsb+\cos kx=\frac{\sin\left(k+\frac12\right)x}{2\sin\frac12x}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\frac12+\cos x+\cos 2x+\dotsb+\cos kx+\cos (k+1)x=\frac{\sin\left(k+1+\frac12\right)x}{2\sin\frac12x}$
$LHS=\frac12+\cos x+\cos 2x+\dotsb+\cos kx+\cos (k+1)x$
$=[\frac12+\cos x+\cos 2x+\dotsb+\cos kx]\,+\cos (k+1)x$
$=[\frac12+\cos x+\cos 2x+\dotsb+\cos kx]\,+\cos (k+1)x$
$=\frac{\sin\left(k+\frac12\right)x}{2\sin\frac12x}+\cos (k+1)x$ (using Step 2)
$=\frac{\sin\left(k+\frac12\right)x}{2\sin\frac12x}+\frac{2\cos (k+1)x\sin\frac12x}{2\sin\frac12x}$
$=\frac{\sin\left(k+\frac12\right)x+2\cos (k+1)x\sin\frac12x}{2\sin\frac12x}$
$=\frac{\sin\left(k+\frac12\right)x+\sin\left(k+\frac32\right)x-\sin\left(k-\frac12\right)x}{2\sin\frac12x}$
$=\frac{\sin\left(k+1+\frac12\right)x}{2\sin\frac12x}$
$=RHS$
Step 4 - Conclusion
If true for $n=k$, then we have proven it is true for $n=k+1$.
Also, as true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
s
Answer and Solution are the same for proofs
To prove: $\frac12+\cos x+\cos 2x+\dotsb+\cos nx=\frac{\sin\left(n+\frac12\right)x}{2\sin\frac12x}$
Step 1 - Prove true for $n=1$
$LHS=\frac12+\cos x$
$RHS=\frac{\sin\left(1+\frac12\right)x}{2\sin\frac12x}$
$=\frac{\sin\frac32x}{2\sin\frac12x}$
$=\frac{3\sin\frac12x-4\sin^3\frac12x}{2\sin\frac12x}$
$=\frac{3-4\sin^2\frac12x}{2}$
$=\frac{3-2(1-\cos x)}{2}$
$=\frac{1+2\cos x}{2}$
$=\frac12+\cos x$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\frac12+\cos x+\cos 2x+\dotsb+\cos kx=\frac{\sin\left(k+\frac12\right)x}{2\sin\frac12x}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\frac12+\cos x+\cos 2x+\dotsb+\cos kx+\cos (k+1)x=\frac{\sin\left(k+1+\frac12\right)x}{2\sin\frac12x}$
$LHS=\frac12+\cos x+\cos 2x+\dotsb+\cos kx+\cos (k+1)x$
$=[\frac12+\cos x+\cos 2x+\dotsb+\cos kx]\,+\cos (k+1)x$
$=[\frac12+\cos x+\cos 2x+\dotsb+\cos kx]\,+\cos (k+1)x$
$=\frac{\sin\left(k+\frac12\right)x}{2\sin\frac12x}+\cos (k+1)x$ (using Step 2)
$=\frac{\sin\left(k+\frac12\right)x}{2\sin\frac12x}+\frac{2\cos (k+1)x\sin\frac12x}{2\sin\frac12x}$
$=\frac{\sin\left(k+\frac12\right)x+2\cos (k+1)x\sin\frac12x}{2\sin\frac12x}$
$=\frac{\sin\left(k+\frac12\right)x+\sin\left(k+\frac32\right)x-\sin\left(k-\frac12\right)x}{2\sin\frac12x}$
$=\frac{\sin\left(k+1+\frac12\right)x}{2\sin\frac12x}$
$=RHS$
Step 4 - Conclusion
If true for $n=k$, then we have proven it is true for $n=k+1$.
Also, as true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
v
To prove: $\frac12+\cos x+\cos 2x+\dotsb+\cos nx=\frac{\sin\left(n+\frac12\right)x}{2\sin\frac12x}$
Step 1 - Prove true for $n=1$
$LHS=\frac12+\cos x$
$RHS=\frac{\sin\left(1+\frac12\right)x}{2\sin\frac12x}$
$=\frac{\sin\frac32x}{2\sin\frac12x}$
$=\frac{3\sin\frac12x-4\sin^3\frac12x}{2\sin\frac12x}$
$=\frac{3-4\sin^2\frac12x}{2}$
$=\frac{3-2(1-\cos x)}{2}$
$=\frac{1+2\cos x}{2}$
$=\frac12+\cos x$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\frac12+\cos x+\cos 2x+\dotsb+\cos kx=\frac{\sin\left(k+\frac12\right)x}{2\sin\frac12x}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\frac12+\cos x+\cos 2x+\dotsb+\cos kx+\cos (k+1)x=\frac{\sin\left(k+1+\frac12\right)x}{2\sin\frac12x}$
$LHS=\frac12+\cos x+\cos 2x+\dotsb+\cos kx+\cos (k+1)x$
$=[\frac12+\cos x+\cos 2x+\dotsb+\cos kx]\,+\cos (k+1)x$
$=[\frac12+\cos x+\cos 2x+\dotsb+\cos kx]\,+\cos (k+1)x$
$=\frac{\sin\left(k+\frac12\right)x}{2\sin\frac12x}+\cos (k+1)x$ (using Step 2)
$=\frac{\sin\left(k+\frac12\right)x}{2\sin\frac12x}+\frac{2\cos (k+1)x\sin\frac12x}{2\sin\frac12x}$
$=\frac{\sin\left(k+\frac12\right)x+2\cos (k+1)x\sin\frac12x}{2\sin\frac12x}$
$=\frac{\sin\left(k+\frac12\right)x+\sin\left(k+\frac32\right)x-\sin\left(k-\frac12\right)x}{2\sin\frac12x}$
$=\frac{\sin\left(k+1+\frac12\right)x}{2\sin\frac12x}$
$=RHS$
Step 4 - Conclusion
If true for $n=k$, then we have proven it is true for $n=k+1$.
Also, as true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
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To prove: $\frac12+\cos x+\cos 2x+\dotsb+\cos nx=\frac{\sin\left(n+\frac12\right)x}{2\sin\frac12x}$
Step 1 - Prove true for $n=1$
$LHS=\frac12+\cos x$
$RHS=\frac{\sin\left(1+\frac12\right)x}{2\sin\frac12x}$
$=\frac{\sin\frac32x}{2\sin\frac12x}$
$=\frac{3\sin\frac12x-4\sin^3\frac12x}{2\sin\frac12x}$
$=\frac{3-4\sin^2\frac12x}{2}$
$=\frac{3-2(1-\cos x)}{2}$
$=\frac{1+2\cos x}{2}$
$=\frac12+\cos x$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\frac12+\cos x+\cos 2x+\dotsb+\cos kx=\frac{\sin\left(k+\frac12\right)x}{2\sin\frac12x}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\frac12+\cos x+\cos 2x+\dotsb+\cos kx+\cos (k+1)x=\frac{\sin\left(k+1+\frac12\right)x}{2\sin\frac12x}$
$LHS=\frac12+\cos x+\cos 2x+\dotsb+\cos kx+\cos (k+1)x$
$=[\frac12+\cos x+\cos 2x+\dotsb+\cos kx]\,+\cos (k+1)x$
$=[\frac12+\cos x+\cos 2x+\dotsb+\cos kx]\,+\cos (k+1)x$
$=\frac{\sin\left(k+\frac12\right)x}{2\sin\frac12x}+\cos (k+1)x$ (using Step 2)
$=\frac{\sin\left(k+\frac12\right)x}{2\sin\frac12x}+\frac{2\cos (k+1)x\sin\frac12x}{2\sin\frac12x}$
$=\frac{\sin\left(k+\frac12\right)x+2\cos (k+1)x\sin\frac12x}{2\sin\frac12x}$
$=\frac{\sin\left(k+\frac12\right)x+\sin\left(k+\frac32\right)x-\sin\left(k-\frac12\right)x}{2\sin\frac12x}$
$=\frac{\sin\left(k+1+\frac12\right)x}{2\sin\frac12x}$
$=RHS$
Step 4 - Conclusion
If true for $n=k$, then we have proven it is true for $n=k+1$.
Also, as true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
d) $\sin x+\sin 2x+\sin 3x+\dots+\sin nx=\frac{\sin\frac12(n+1)x\sin\frac12nx}{\sin\frac12x}$
a
Answer and Solution are the same for proofs
To prove: $\sin x+\sin 2x+\sin 3x+\dots+\sin nx=\frac{\sin\frac12(n+1)x\sin\frac12nx}{\sin\frac12x}$ for all positive integers, $n$
Step 1 - Prove true for $n=1$
$LHS=\sin x$
$RHS=\frac{\sin\frac12\times2\times x\times\sin\frac12\times1\times x}{\sin\frac12x}$
$=\frac{\sin x\sin\frac12x}{\sin\frac12x}$
$=\sin x$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\sin x+\sin 2x+\sin 3x+\dots+\sin kx=\frac{\sin\frac12(k+1)x\sin\frac12kx}{\sin\frac12x}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\sin x+\sin 2x+\sin 3x+\dots+\sin kx+\sin(k+1)x=\frac{\sin\frac12(k+2)x\sin\frac12(k+1)x}{\sin\frac12x}$
$LHS=\sin x+\sin 2x+\sin 3x+\dots+\sin kx+\sin(k+1)x$
$=\frac{\sin\frac12(k+1)x\sin\frac12kx}{\sin\frac12x}+\sin(k+1)x$ (using Step 2)
$=\frac{\sin\frac12(k+1)x\sin\frac12kx+\sin(k+1)x\sin\frac12x}{\sin\frac12x}$
Aside:
Use $\sin A\sin B=\frac12\left(\cos(A-B)-\cos(A+B)\right)$
so $=\frac{1}{\sin\frac12 x}\times\left[\frac12\left(\cos\frac12x-\cos(k+\frac12)x\right)+\frac12\left(\cos(k+\frac12)x-\cos(k+\frac32)x\right)\right]$
$=\frac{1}{\sin\frac12 x}\times\frac12\left(\cos\frac12x-\cos(k+\frac32)x\right)$
$=\frac{1}{\sin\frac12 x}\times\sin\frac{(k+1)x}{2}\sin\frac{(k+2)x}{2}$
$=RHS$
Step 4 - Conclusion
If true for $n=k$, then we have proven it is true for $n=k+1$.
Also, as true when $n=1$, it is also true for all positive integers by mathematical induction.
s
Answer and Solution are the same for proofs
To prove: $\sin x+\sin 2x+\sin 3x+\dots+\sin nx=\frac{\sin\frac12(n+1)x\sin\frac12nx}{\sin\frac12x}$ for all positive integers, $n$
Step 1 - Prove true for $n=1$
$LHS=\sin x$
$RHS=\frac{\sin\frac12\times2\times x\times\sin\frac12\times1\times x}{\sin\frac12x}$
$=\frac{\sin x\sin\frac12x}{\sin\frac12x}$
$=\sin x$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\sin x+\sin 2x+\sin 3x+\dots+\sin kx=\frac{\sin\frac12(k+1)x\sin\frac12kx}{\sin\frac12x}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\sin x+\sin 2x+\sin 3x+\dots+\sin kx+\sin(k+1)x=\frac{\sin\frac12(k+2)x\sin\frac12(k+1)x}{\sin\frac12x}$
$LHS=\sin x+\sin 2x+\sin 3x+\dots+\sin kx+\sin(k+1)x$
$=\frac{\sin\frac12(k+1)x\sin\frac12kx}{\sin\frac12x}+\sin(k+1)x$ (using Step 2)
$=\frac{\sin\frac12(k+1)x\sin\frac12kx+\sin(k+1)x\sin\frac12x}{\sin\frac12x}$
Aside:
Use $\sin A\sin B=\frac12\left(\cos(A-B)-\cos(A+B)\right)$
so $=\frac{1}{\sin\frac12 x}\times\left[\frac12\left(\cos\frac12x-\cos(k+\frac12)x\right)+\frac12\left(\cos(k+\frac12)x-\cos(k+\frac32)x\right)\right]$
$=\frac{1}{\sin\frac12 x}\times\frac12\left(\cos\frac12x-\cos(k+\frac32)x\right)$
$=\frac{1}{\sin\frac12 x}\times\sin\frac{(k+1)x}{2}\sin\frac{(k+2)x}{2}$
$=RHS$
Step 4 - Conclusion
If true for $n=k$, then we have proven it is true for $n=k+1$.
Also, as true when $n=1$, it is also true for all positive integers by mathematical induction.
v
To prove: $\sin x+\sin 2x+\sin 3x+\dots+\sin nx=\frac{\sin\frac12(n+1)x\sin\frac12nx}{\sin\frac12x}$ for all positive integers, $n$
Step 1 - Prove true for $n=1$
$LHS=\sin x$
$RHS=\frac{\sin\frac12\times2\times x\times\sin\frac12\times1\times x}{\sin\frac12x}$
$=\frac{\sin x\sin\frac12x}{\sin\frac12x}$
$=\sin x$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\sin x+\sin 2x+\sin 3x+\dots+\sin kx=\frac{\sin\frac12(k+1)x\sin\frac12kx}{\sin\frac12x}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\sin x+\sin 2x+\sin 3x+\dots+\sin kx+\sin(k+1)x=\frac{\sin\frac12(k+2)x\sin\frac12(k+1)x}{\sin\frac12x}$
$LHS=\sin x+\sin 2x+\sin 3x+\dots+\sin kx+\sin(k+1)x$
$=\frac{\sin\frac12(k+1)x\sin\frac12kx}{\sin\frac12x}+\sin(k+1)x$ (using Step 2)
$=\frac{\sin\frac12(k+1)x\sin\frac12kx+\sin(k+1)x\sin\frac12x}{\sin\frac12x}$
Aside:
Use $\sin A\sin B=\frac12\left(\cos(A-B)-\cos(A+B)\right)$
so $=\frac{1}{\sin\frac12 x}\times\left[\frac12\left(\cos\frac12x-\cos(k+\frac12)x\right)+\frac12\left(\cos(k+\frac12)x-\cos(k+\frac32)x\right)\right]$
$=\frac{1}{\sin\frac12 x}\times\frac12\left(\cos\frac12x-\cos(k+\frac32)x\right)$
$=\frac{1}{\sin\frac12 x}\times\sin\frac{(k+1)x}{2}\sin\frac{(k+2)x}{2}$
$=RHS$
Step 4 - Conclusion
If true for $n=k$, then we have proven it is true for $n=k+1$.
Also, as true when $n=1$, it is also true for all positive integers by mathematical induction.
Question ID: 10050070011
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Answer by: ada
To prove: $\sin x+\sin 2x+\sin 3x+\dots+\sin nx=\frac{\sin\frac12(n+1)x\sin\frac12nx}{\sin\frac12x}$ for all positive integers, $n$
Step 1 - Prove true for $n=1$
$LHS=\sin x$
$RHS=\frac{\sin\frac12\times2\times x\times\sin\frac12\times1\times x}{\sin\frac12x}$
$=\frac{\sin x\sin\frac12x}{\sin\frac12x}$
$=\sin x$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\sin x+\sin 2x+\sin 3x+\dots+\sin kx=\frac{\sin\frac12(k+1)x\sin\frac12kx}{\sin\frac12x}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\sin x+\sin 2x+\sin 3x+\dots+\sin kx+\sin(k+1)x=\frac{\sin\frac12(k+2)x\sin\frac12(k+1)x}{\sin\frac12x}$
$LHS=\sin x+\sin 2x+\sin 3x+\dots+\sin kx+\sin(k+1)x$
$=\frac{\sin\frac12(k+1)x\sin\frac12kx}{\sin\frac12x}+\sin(k+1)x$ (using Step 2)
$=\frac{\sin\frac12(k+1)x\sin\frac12kx+\sin(k+1)x\sin\frac12x}{\sin\frac12x}$
Aside:
Use $\sin A\sin B=\frac12\left(\cos(A-B)-\cos(A+B)\right)$
so $=\frac{1}{\sin\frac12 x}\times\left[\frac12\left(\cos\frac12x-\cos(k+\frac12)x\right)+\frac12\left(\cos(k+\frac12)x-\cos(k+\frac32)x\right)\right]$
$=\frac{1}{\sin\frac12 x}\times\frac12\left(\cos\frac12x-\cos(k+\frac32)x\right)$
$=\frac{1}{\sin\frac12 x}\times\sin\frac{(k+1)x}{2}\sin\frac{(k+2)x}{2}$
$=RHS$
Step 4 - Conclusion
If true for $n=k$, then we have proven it is true for $n=k+1$.
Also, as true when $n=1$, it is also true for all positive integers by mathematical induction.
Question 2
(Requires knowledge of complex numbers). Prove by mathematical induction: ( where $i=\sqrt{-1}$)
a) $(\cos(x)+i\sin(x))^n=\cos(nx)+i\sin(nx)$ for $n\ge0$ and $n\in\mathbb{Z}^+$
a
Answer and Solution are the same for proofs
To prove: $(\cos(x)+i\sin(x))^n=\cos(nx)+i\sin(nx)$ for $n\ge0$ and $n\in\mathbb{Z}^+$:
Step 1 - Prove true for $n=0$
$LHS=(\cos(x)+i\sin(x))^0$
$=1$
$RHS=\cos(0)+i\sin(0)$
$=1+0$
$=1$
$=LHS$
So true for $n=0$
Step 2 - Assume true for $n=k$
$(\cos(x)+i\sin(x))^k=\cos(kx)+i\sin(kx)$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$(\cos(x)+i\sin(x))^{k+1}=\cos((k+1)x)+i\sin((k+1)x)$
$LHS=(\cos(x)+i\sin(x))^{k+1}$
$=(\cos(x)+i\sin(x))^k((\cos(x)+i\sin(x))$
$=(\cos(kx)+i\sin(kx))((\cos(x)+i\sin(x))$ (using Step 2)
$=\cos(kx)\cos(x)+i\cos(kx)\sin(x)+i\cos(x)\sin(kx)-\sin(kx)\sin(x)$
$=\cos(kx)\cos(x)-\sin(kx)\sin(x)+i(\cos(kx)\sin(x)+\cos(x)\sin(kx))$
$=\cos(kx+x)+i\sin(kx+x)$
$=\cos((k+1)x)+i\sin((k+1)x)$
$=RHS$
Step 4 - Conclusion
If true for $n=k$, then we have proven it is true for $n=k+1$.
Also, as true when $n=0$, it is also true for all $n\ge0$ and $n\in\mathbb{Z}^+$ by mathematical induction.
s
Answer and Solution are the same for proofs
To prove: $(\cos(x)+i\sin(x))^n=\cos(nx)+i\sin(nx)$ for $n\ge0$ and $n\in\mathbb{Z}^+$:
Step 1 - Prove true for $n=0$
$LHS=(\cos(x)+i\sin(x))^0$
$=1$
$RHS=\cos(0)+i\sin(0)$
$=1+0$
$=1$
$=LHS$
So true for $n=0$
Step 2 - Assume true for $n=k$
$(\cos(x)+i\sin(x))^k=\cos(kx)+i\sin(kx)$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$(\cos(x)+i\sin(x))^{k+1}=\cos((k+1)x)+i\sin((k+1)x)$
$LHS=(\cos(x)+i\sin(x))^{k+1}$
$=(\cos(x)+i\sin(x))^k((\cos(x)+i\sin(x))$
$=(\cos(kx)+i\sin(kx))((\cos(x)+i\sin(x))$ (using Step 2)
$=\cos(kx)\cos(x)+i\cos(kx)\sin(x)+i\cos(x)\sin(kx)-\sin(kx)\sin(x)$
$=\cos(kx)\cos(x)-\sin(kx)\sin(x)+i(\cos(kx)\sin(x)+\cos(x)\sin(kx))$
$=\cos(kx+x)+i\sin(kx+x)$
$=\cos((k+1)x)+i\sin((k+1)x)$
$=RHS$
Step 4 - Conclusion
If true for $n=k$, then we have proven it is true for $n=k+1$.
Also, as true when $n=0$, it is also true for all $n\ge0$ and $n\in\mathbb{Z}^+$ by mathematical induction.
v
To prove: $(\cos(x)+i\sin(x))^n=\cos(nx)+i\sin(nx)$ for $n\ge0$ and $n\in\mathbb{Z}^+$:
Step 1 - Prove true for $n=0$
$LHS=(\cos(x)+i\sin(x))^0$
$=1$
$RHS=\cos(0)+i\sin(0)$
$=1+0$
$=1$
$=LHS$
So true for $n=0$
Step 2 - Assume true for $n=k$
$(\cos(x)+i\sin(x))^k=\cos(kx)+i\sin(kx)$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$(\cos(x)+i\sin(x))^{k+1}=\cos((k+1)x)+i\sin((k+1)x)$
$LHS=(\cos(x)+i\sin(x))^{k+1}$
$=(\cos(x)+i\sin(x))^k((\cos(x)+i\sin(x))$
$=(\cos(kx)+i\sin(kx))((\cos(x)+i\sin(x))$ (using Step 2)
$=\cos(kx)\cos(x)+i\cos(kx)\sin(x)+i\cos(x)\sin(kx)-\sin(kx)\sin(x)$
$=\cos(kx)\cos(x)-\sin(kx)\sin(x)+i(\cos(kx)\sin(x)+\cos(x)\sin(kx))$
$=\cos(kx+x)+i\sin(kx+x)$
$=\cos((k+1)x)+i\sin((k+1)x)$
$=RHS$
Step 4 - Conclusion
If true for $n=k$, then we have proven it is true for $n=k+1$.
Also, as true when $n=0$, it is also true for all $n\ge0$ and $n\in\mathbb{Z}^+$ by mathematical induction.
Question ID: 10050080010
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To prove: $(\cos(x)+i\sin(x))^n=\cos(nx)+i\sin(nx)$ for $n\ge0$ and $n\in\mathbb{Z}^+$:
Step 1 - Prove true for $n=0$
$LHS=(\cos(x)+i\sin(x))^0$
$=1$
$RHS=\cos(0)+i\sin(0)$
$=1+0$
$=1$
$=LHS$
So true for $n=0$
Step 2 - Assume true for $n=k$
$(\cos(x)+i\sin(x))^k=\cos(kx)+i\sin(kx)$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$(\cos(x)+i\sin(x))^{k+1}=\cos((k+1)x)+i\sin((k+1)x)$
$LHS=(\cos(x)+i\sin(x))^{k+1}$
$=(\cos(x)+i\sin(x))^k((\cos(x)+i\sin(x))$
$=(\cos(kx)+i\sin(kx))((\cos(x)+i\sin(x))$ (using Step 2)
$=\cos(kx)\cos(x)+i\cos(kx)\sin(x)+i\cos(x)\sin(kx)-\sin(kx)\sin(x)$
$=\cos(kx)\cos(x)-\sin(kx)\sin(x)+i(\cos(kx)\sin(x)+\cos(x)\sin(kx))$
$=\cos(kx+x)+i\sin(kx+x)$
$=\cos((k+1)x)+i\sin((k+1)x)$
$=RHS$
Step 4 - Conclusion
If true for $n=k$, then we have proven it is true for $n=k+1$.
Also, as true when $n=0$, it is also true for all $n\ge0$ and $n\in\mathbb{Z}^+$ by mathematical induction.
Question 1
Prove by mathematical induction for $n\ge1$ and $n\in\mathbb{Z}^+$:
a) $\cos x\times\cos 2x\times\dots\times\cos(2^{n-1}x)=\frac{\sin(2^nx)}{2^n\sin(x)}$
b) $\cos\left(\frac{x}{2}\right) + \cos\left(\frac{3x}{2}\right)+\cdots + \cos\left(\frac{(2n-1)x}{2}\right) = \frac{\sin nx}{2\sin\left(\frac x 2\right)}$
c) $\frac12+\cos x+\cos 2x+\dotsb+\cos nx=\frac{\sin\left(n+\frac12\right)x}{2\sin\frac12x}$
d) $\sin x+\sin 2x+\sin 3x+\dots+\sin nx=\frac{\sin\frac12(n+1)x\sin\frac12nx}{\sin\frac12x}$
Question 2
(Requires knowledge of complex numbers). Prove by mathematical induction: ( where $i=\sqrt{-1}$)
a) $(\cos(x)+i\sin(x))^n=\cos(nx)+i\sin(nx)$ for $n\ge0$ and $n\in\mathbb{Z}^+$
Answers
Question 1
a) Answer and Solution are the same for proofs
To prove: $\cos x\times\cos 2x\times\dots\times\cos(2^{n-1}x)=\frac{\sin(2^nx)}{2^n\sin(x)}$ for $n\ge1$ and $n\in\mathbb{Z}^+$:
Step 1 - Prove true for $n=1$
$LHS=\cos x$
$RHS=\frac{\sin(2x)}{2\sin x}$
$=\frac{2\sin x\cos x}{2\sin x}$
$=\cos x$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\cos x\times\cos 2x\times\dots\times\cos(2^{k-1}x)=\frac{\sin(2^kx)}{2^k\sin(x)}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\cos x\times\cos 2x\times\dots\times\cos(2^{(k+1)-1}x)=\frac{\sin(2^{k+1}x)}{2^{k+1}\sin(x)}$
$LHS=\cos x\times\cos 2x\times\dots\times\cos(2^{k-1}x)\times\cos(2^kx)$
$=\frac{\sin(2^kx)}{2^k\sin(x)}\times\cos(2^kx)$ (using Step 2)
$=\frac{\sin(2^kx)\cos(2^kx)}{2^k\sin(x)}$
$=\frac{2\sin(2^kx)\cos(2^kx)}{2^{k+1}\sin(x)}$
$=\frac{\sin(2\times2^kx)}{2^{k+1}\sin(x)}$
$=\frac{\sin(2^{k+1}x)}{2^{k+1}\sin(x)}$
$=RHS$
Step 4 - Conclusion
If true for $n=k$, then we have proven it is true for $n=k+1$.
Also, as true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
To prove: $\cos x\times\cos 2x\times\dots\times\cos(2^{n-1}x)=\frac{\sin(2^nx)}{2^n\sin(x)}$ for $n\ge1$ and $n\in\mathbb{Z}^+$:
Step 1 - Prove true for $n=1$
$LHS=\cos x$
$RHS=\frac{\sin(2x)}{2\sin x}$
$=\frac{2\sin x\cos x}{2\sin x}$
$=\cos x$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\cos x\times\cos 2x\times\dots\times\cos(2^{k-1}x)=\frac{\sin(2^kx)}{2^k\sin(x)}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\cos x\times\cos 2x\times\dots\times\cos(2^{(k+1)-1}x)=\frac{\sin(2^{k+1}x)}{2^{k+1}\sin(x)}$
$LHS=\cos x\times\cos 2x\times\dots\times\cos(2^{k-1}x)\times\cos(2^kx)$
$=\frac{\sin(2^kx)}{2^k\sin(x)}\times\cos(2^kx)$ (using Step 2)
$=\frac{\sin(2^kx)\cos(2^kx)}{2^k\sin(x)}$
$=\frac{2\sin(2^kx)\cos(2^kx)}{2^{k+1}\sin(x)}$
$=\frac{\sin(2\times2^kx)}{2^{k+1}\sin(x)}$
$=\frac{\sin(2^{k+1}x)}{2^{k+1}\sin(x)}$
$=RHS$
Step 4 - Conclusion
If true for $n=k$, then we have proven it is true for $n=k+1$.
Also, as true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
b) Answer and Solution are the same for proofs
To prove: $\cos\left(\frac{x}{2}\right) + \cos\left(\frac{3x}{2}\right)+\cdots + \cos\left(\frac{(2n-1)x}{2}\right) = \frac{\sin nx}{2\sin\left(\frac x 2\right)}$ for $n\ge1$ and $n\in\mathbb{Z}^+$:
Step 1 - Prove true for $n=1$
$LHS=\cos\left(\frac x2\right)$
$RHS=\frac{\sin x}{2\sin\left(\frac x 2\right)}$
$=\frac{2\sin\left(\frac x2\right)\cos\left(\frac x2\right)}{2\sin\left(\frac x 2\right)}$
$=\cos\left(\frac x2\right)$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\cos\left(\frac{x}{2}\right) + \cos\left(\frac{3x}{2}\right)+\cdots + \cos\left(\frac{(2k-1)x}{2}\right) = \frac{\sin kx}{2\sin\left(\frac x 2\right)}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\cos\left(\frac{x}{2}\right) + \cos\left(\frac{3x}{2}\right)+\cdots + \cos\left(\frac{(2k-1)x}{2}\right) + \cos\left(\frac{(2(k+1)-1)x}{2}\right) = \frac{\sin(k+1)x}{2\sin\left(\frac x 2\right)}$
$LHS=\cos\left(\frac{x}{2}\right) + \cos\left(\frac{3x}{2}\right)+\cdots + \cos\left(\frac{(2k-1)x}{2}\right) + \cos\left(\frac{(2k+1)x}{2}\right)$
$=\frac{\sin kx}{2\sin\left(\frac x 2\right)}+ \cos\left(\frac{(2k+1)x}{2}\right)$ (using Step 2)
$=\frac{\sin kx + 2\sin\left(\frac x 2\right)\cos\left(kx+\frac x 2\right)}{2\sin\left(\frac x 2\right)}$
$=\frac{\sin kx + 2\sin\left(\frac x 2\right)\left[\cos kx\cos\left(\frac x 2\right)-\sin kx\sin\left(\frac x 2\right)\right]}{2\sin\left(\frac x 2\right)}$
$\frac{\sin kx \left[1-2\sin^2\left(\frac x 2\right)\right]+2\sin\left(\frac x 2\right)\cos\left(\frac x 2\right)\cos kx}{2\sin\left(\frac x 2\right)}$
$=\frac{\sin kx\cos x+\sin x\cos kx}{2\sin\left(\frac x 2\right)}$
$=\frac{\sin(kx+x)}{2\sin\left(\frac x 2\right)}$
$=\frac{\sin(k+1)x}{2\sin\left(\frac x 2\right)}$
$=RHS$
Step 4 - Conclusion
If true for $n=k$, then we have proven it is true for $n=k+1$.
Also, as true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
To prove: $\cos\left(\frac{x}{2}\right) + \cos\left(\frac{3x}{2}\right)+\cdots + \cos\left(\frac{(2n-1)x}{2}\right) = \frac{\sin nx}{2\sin\left(\frac x 2\right)}$ for $n\ge1$ and $n\in\mathbb{Z}^+$:
Step 1 - Prove true for $n=1$
$LHS=\cos\left(\frac x2\right)$
$RHS=\frac{\sin x}{2\sin\left(\frac x 2\right)}$
$=\frac{2\sin\left(\frac x2\right)\cos\left(\frac x2\right)}{2\sin\left(\frac x 2\right)}$
$=\cos\left(\frac x2\right)$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\cos\left(\frac{x}{2}\right) + \cos\left(\frac{3x}{2}\right)+\cdots + \cos\left(\frac{(2k-1)x}{2}\right) = \frac{\sin kx}{2\sin\left(\frac x 2\right)}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\cos\left(\frac{x}{2}\right) + \cos\left(\frac{3x}{2}\right)+\cdots + \cos\left(\frac{(2k-1)x}{2}\right) + \cos\left(\frac{(2(k+1)-1)x}{2}\right) = \frac{\sin(k+1)x}{2\sin\left(\frac x 2\right)}$
$LHS=\cos\left(\frac{x}{2}\right) + \cos\left(\frac{3x}{2}\right)+\cdots + \cos\left(\frac{(2k-1)x}{2}\right) + \cos\left(\frac{(2k+1)x}{2}\right)$
$=\frac{\sin kx}{2\sin\left(\frac x 2\right)}+ \cos\left(\frac{(2k+1)x}{2}\right)$ (using Step 2)
$=\frac{\sin kx + 2\sin\left(\frac x 2\right)\cos\left(kx+\frac x 2\right)}{2\sin\left(\frac x 2\right)}$
$=\frac{\sin kx + 2\sin\left(\frac x 2\right)\left[\cos kx\cos\left(\frac x 2\right)-\sin kx\sin\left(\frac x 2\right)\right]}{2\sin\left(\frac x 2\right)}$
$\frac{\sin kx \left[1-2\sin^2\left(\frac x 2\right)\right]+2\sin\left(\frac x 2\right)\cos\left(\frac x 2\right)\cos kx}{2\sin\left(\frac x 2\right)}$
$=\frac{\sin kx\cos x+\sin x\cos kx}{2\sin\left(\frac x 2\right)}$
$=\frac{\sin(kx+x)}{2\sin\left(\frac x 2\right)}$
$=\frac{\sin(k+1)x}{2\sin\left(\frac x 2\right)}$
$=RHS$
Step 4 - Conclusion
If true for $n=k$, then we have proven it is true for $n=k+1$.
Also, as true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
c) Answer and Solution are the same for proofs
To prove: $\frac12+\cos x+\cos 2x+\dotsb+\cos nx=\frac{\sin\left(n+\frac12\right)x}{2\sin\frac12x}$
Step 1 - Prove true for $n=1$
$LHS=\frac12+\cos x$
$RHS=\frac{\sin\left(1+\frac12\right)x}{2\sin\frac12x}$
$=\frac{\sin\frac32x}{2\sin\frac12x}$
$=\frac{3\sin\frac12x-4\sin^3\frac12x}{2\sin\frac12x}$
$=\frac{3-4\sin^2\frac12x}{2}$
$=\frac{3-2(1-\cos x)}{2}$
$=\frac{1+2\cos x}{2}$
$=\frac12+\cos x$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\frac12+\cos x+\cos 2x+\dotsb+\cos kx=\frac{\sin\left(k+\frac12\right)x}{2\sin\frac12x}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\frac12+\cos x+\cos 2x+\dotsb+\cos kx+\cos (k+1)x=\frac{\sin\left(k+1+\frac12\right)x}{2\sin\frac12x}$
$LHS=\frac12+\cos x+\cos 2x+\dotsb+\cos kx+\cos (k+1)x$
$=[\frac12+\cos x+\cos 2x+\dotsb+\cos kx]\,+\cos (k+1)x$
$=[\frac12+\cos x+\cos 2x+\dotsb+\cos kx]\,+\cos (k+1)x$
$=\frac{\sin\left(k+\frac12\right)x}{2\sin\frac12x}+\cos (k+1)x$ (using Step 2)
$=\frac{\sin\left(k+\frac12\right)x}{2\sin\frac12x}+\frac{2\cos (k+1)x\sin\frac12x}{2\sin\frac12x}$
$=\frac{\sin\left(k+\frac12\right)x+2\cos (k+1)x\sin\frac12x}{2\sin\frac12x}$
$=\frac{\sin\left(k+\frac12\right)x+\sin\left(k+\frac32\right)x-\sin\left(k-\frac12\right)x}{2\sin\frac12x}$
$=\frac{\sin\left(k+1+\frac12\right)x}{2\sin\frac12x}$
$=RHS$
Step 4 - Conclusion
If true for $n=k$, then we have proven it is true for $n=k+1$.
Also, as true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
To prove: $\frac12+\cos x+\cos 2x+\dotsb+\cos nx=\frac{\sin\left(n+\frac12\right)x}{2\sin\frac12x}$
Step 1 - Prove true for $n=1$
$LHS=\frac12+\cos x$
$RHS=\frac{\sin\left(1+\frac12\right)x}{2\sin\frac12x}$
$=\frac{\sin\frac32x}{2\sin\frac12x}$
$=\frac{3\sin\frac12x-4\sin^3\frac12x}{2\sin\frac12x}$
$=\frac{3-4\sin^2\frac12x}{2}$
$=\frac{3-2(1-\cos x)}{2}$
$=\frac{1+2\cos x}{2}$
$=\frac12+\cos x$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\frac12+\cos x+\cos 2x+\dotsb+\cos kx=\frac{\sin\left(k+\frac12\right)x}{2\sin\frac12x}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\frac12+\cos x+\cos 2x+\dotsb+\cos kx+\cos (k+1)x=\frac{\sin\left(k+1+\frac12\right)x}{2\sin\frac12x}$
$LHS=\frac12+\cos x+\cos 2x+\dotsb+\cos kx+\cos (k+1)x$
$=[\frac12+\cos x+\cos 2x+\dotsb+\cos kx]\,+\cos (k+1)x$
$=[\frac12+\cos x+\cos 2x+\dotsb+\cos kx]\,+\cos (k+1)x$
$=\frac{\sin\left(k+\frac12\right)x}{2\sin\frac12x}+\cos (k+1)x$ (using Step 2)
$=\frac{\sin\left(k+\frac12\right)x}{2\sin\frac12x}+\frac{2\cos (k+1)x\sin\frac12x}{2\sin\frac12x}$
$=\frac{\sin\left(k+\frac12\right)x+2\cos (k+1)x\sin\frac12x}{2\sin\frac12x}$
$=\frac{\sin\left(k+\frac12\right)x+\sin\left(k+\frac32\right)x-\sin\left(k-\frac12\right)x}{2\sin\frac12x}$
$=\frac{\sin\left(k+1+\frac12\right)x}{2\sin\frac12x}$
$=RHS$
Step 4 - Conclusion
If true for $n=k$, then we have proven it is true for $n=k+1$.
Also, as true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
d) Answer and Solution are the same for proofs
To prove: $\sin x+\sin 2x+\sin 3x+\dots+\sin nx=\frac{\sin\frac12(n+1)x\sin\frac12nx}{\sin\frac12x}$ for all positive integers, $n$
Step 1 - Prove true for $n=1$
$LHS=\sin x$
$RHS=\frac{\sin\frac12\times2\times x\times\sin\frac12\times1\times x}{\sin\frac12x}$
$=\frac{\sin x\sin\frac12x}{\sin\frac12x}$
$=\sin x$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\sin x+\sin 2x+\sin 3x+\dots+\sin kx=\frac{\sin\frac12(k+1)x\sin\frac12kx}{\sin\frac12x}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\sin x+\sin 2x+\sin 3x+\dots+\sin kx+\sin(k+1)x=\frac{\sin\frac12(k+2)x\sin\frac12(k+1)x}{\sin\frac12x}$
$LHS=\sin x+\sin 2x+\sin 3x+\dots+\sin kx+\sin(k+1)x$
$=\frac{\sin\frac12(k+1)x\sin\frac12kx}{\sin\frac12x}+\sin(k+1)x$ (using Step 2)
$=\frac{\sin\frac12(k+1)x\sin\frac12kx+\sin(k+1)x\sin\frac12x}{\sin\frac12x}$
Aside:
Use $\sin A\sin B=\frac12\left(\cos(A-B)-\cos(A+B)\right)$
so $=\frac{1}{\sin\frac12 x}\times\left[\frac12\left(\cos\frac12x-\cos(k+\frac12)x\right)+\frac12\left(\cos(k+\frac12)x-\cos(k+\frac32)x\right)\right]$
$=\frac{1}{\sin\frac12 x}\times\frac12\left(\cos\frac12x-\cos(k+\frac32)x\right)$
$=\frac{1}{\sin\frac12 x}\times\sin\frac{(k+1)x}{2}\sin\frac{(k+2)x}{2}$
$=RHS$
Step 4 - Conclusion
If true for $n=k$, then we have proven it is true for $n=k+1$.
Also, as true when $n=1$, it is also true for all positive integers by mathematical induction.
To prove: $\sin x+\sin 2x+\sin 3x+\dots+\sin nx=\frac{\sin\frac12(n+1)x\sin\frac12nx}{\sin\frac12x}$ for all positive integers, $n$
Step 1 - Prove true for $n=1$
$LHS=\sin x$
$RHS=\frac{\sin\frac12\times2\times x\times\sin\frac12\times1\times x}{\sin\frac12x}$
$=\frac{\sin x\sin\frac12x}{\sin\frac12x}$
$=\sin x$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\sin x+\sin 2x+\sin 3x+\dots+\sin kx=\frac{\sin\frac12(k+1)x\sin\frac12kx}{\sin\frac12x}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\sin x+\sin 2x+\sin 3x+\dots+\sin kx+\sin(k+1)x=\frac{\sin\frac12(k+2)x\sin\frac12(k+1)x}{\sin\frac12x}$
$LHS=\sin x+\sin 2x+\sin 3x+\dots+\sin kx+\sin(k+1)x$
$=\frac{\sin\frac12(k+1)x\sin\frac12kx}{\sin\frac12x}+\sin(k+1)x$ (using Step 2)
$=\frac{\sin\frac12(k+1)x\sin\frac12kx+\sin(k+1)x\sin\frac12x}{\sin\frac12x}$
Aside:
Use $\sin A\sin B=\frac12\left(\cos(A-B)-\cos(A+B)\right)$
so $=\frac{1}{\sin\frac12 x}\times\left[\frac12\left(\cos\frac12x-\cos(k+\frac12)x\right)+\frac12\left(\cos(k+\frac12)x-\cos(k+\frac32)x\right)\right]$
$=\frac{1}{\sin\frac12 x}\times\frac12\left(\cos\frac12x-\cos(k+\frac32)x\right)$
$=\frac{1}{\sin\frac12 x}\times\sin\frac{(k+1)x}{2}\sin\frac{(k+2)x}{2}$
$=RHS$
Step 4 - Conclusion
If true for $n=k$, then we have proven it is true for $n=k+1$.
Also, as true when $n=1$, it is also true for all positive integers by mathematical induction.
Question 2
a) Answer and Solution are the same for proofs
To prove: $(\cos(x)+i\sin(x))^n=\cos(nx)+i\sin(nx)$ for $n\ge0$ and $n\in\mathbb{Z}^+$:
Step 1 - Prove true for $n=0$
$LHS=(\cos(x)+i\sin(x))^0$
$=1$
$RHS=\cos(0)+i\sin(0)$
$=1+0$
$=1$
$=LHS$
So true for $n=0$
Step 2 - Assume true for $n=k$
$(\cos(x)+i\sin(x))^k=\cos(kx)+i\sin(kx)$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$(\cos(x)+i\sin(x))^{k+1}=\cos((k+1)x)+i\sin((k+1)x)$
$LHS=(\cos(x)+i\sin(x))^{k+1}$
$=(\cos(x)+i\sin(x))^k((\cos(x)+i\sin(x))$
$=(\cos(kx)+i\sin(kx))((\cos(x)+i\sin(x))$ (using Step 2)
$=\cos(kx)\cos(x)+i\cos(kx)\sin(x)+i\cos(x)\sin(kx)-\sin(kx)\sin(x)$
$=\cos(kx)\cos(x)-\sin(kx)\sin(x)+i(\cos(kx)\sin(x)+\cos(x)\sin(kx))$
$=\cos(kx+x)+i\sin(kx+x)$
$=\cos((k+1)x)+i\sin((k+1)x)$
$=RHS$
Step 4 - Conclusion
If true for $n=k$, then we have proven it is true for $n=k+1$.
Also, as true when $n=0$, it is also true for all $n\ge0$ and $n\in\mathbb{Z}^+$ by mathematical induction.
To prove: $(\cos(x)+i\sin(x))^n=\cos(nx)+i\sin(nx)$ for $n\ge0$ and $n\in\mathbb{Z}^+$:
Step 1 - Prove true for $n=0$
$LHS=(\cos(x)+i\sin(x))^0$
$=1$
$RHS=\cos(0)+i\sin(0)$
$=1+0$
$=1$
$=LHS$
So true for $n=0$
Step 2 - Assume true for $n=k$
$(\cos(x)+i\sin(x))^k=\cos(kx)+i\sin(kx)$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$(\cos(x)+i\sin(x))^{k+1}=\cos((k+1)x)+i\sin((k+1)x)$
$LHS=(\cos(x)+i\sin(x))^{k+1}$
$=(\cos(x)+i\sin(x))^k((\cos(x)+i\sin(x))$
$=(\cos(kx)+i\sin(kx))((\cos(x)+i\sin(x))$ (using Step 2)
$=\cos(kx)\cos(x)+i\cos(kx)\sin(x)+i\cos(x)\sin(kx)-\sin(kx)\sin(x)$
$=\cos(kx)\cos(x)-\sin(kx)\sin(x)+i(\cos(kx)\sin(x)+\cos(x)\sin(kx))$
$=\cos(kx+x)+i\sin(kx+x)$
$=\cos((k+1)x)+i\sin((k+1)x)$
$=RHS$
Step 4 - Conclusion
If true for $n=k$, then we have proven it is true for $n=k+1$.
Also, as true when $n=0$, it is also true for all $n\ge0$ and $n\in\mathbb{Z}^+$ by mathematical induction.
Question 1
Prove by mathematical induction for $n\ge1$ and $n\in\mathbb{Z}^+$:
a) $\cos x\times\cos 2x\times\dots\times\cos(2^{n-1}x)=\frac{\sin(2^nx)}{2^n\sin(x)}$
b) $\cos\left(\frac{x}{2}\right) + \cos\left(\frac{3x}{2}\right)+\cdots + \cos\left(\frac{(2n-1)x}{2}\right) = \frac{\sin nx}{2\sin\left(\frac x 2\right)}$
c) $\frac12+\cos x+\cos 2x+\dotsb+\cos nx=\frac{\sin\left(n+\frac12\right)x}{2\sin\frac12x}$
d) $\sin x+\sin 2x+\sin 3x+\dots+\sin nx=\frac{\sin\frac12(n+1)x\sin\frac12nx}{\sin\frac12x}$
Question 2
(Requires knowledge of complex numbers). Prove by mathematical induction: ( where $i=\sqrt{-1}$)
a) $(\cos(x)+i\sin(x))^n=\cos(nx)+i\sin(nx)$ for $n\ge0$ and $n\in\mathbb{Z}^+$
Answers
Question 1
a) Answer and Solution are the same for proofs
To prove: $\cos x\times\cos 2x\times\dots\times\cos(2^{n-1}x)=\frac{\sin(2^nx)}{2^n\sin(x)}$ for $n\ge1$ and $n\in\mathbb{Z}^+$:
Step 1 - Prove true for $n=1$
$LHS=\cos x$
$RHS=\frac{\sin(2x)}{2\sin x}$
$=\frac{2\sin x\cos x}{2\sin x}$
$=\cos x$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\cos x\times\cos 2x\times\dots\times\cos(2^{k-1}x)=\frac{\sin(2^kx)}{2^k\sin(x)}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\cos x\times\cos 2x\times\dots\times\cos(2^{(k+1)-1}x)=\frac{\sin(2^{k+1}x)}{2^{k+1}\sin(x)}$
$LHS=\cos x\times\cos 2x\times\dots\times\cos(2^{k-1}x)\times\cos(2^kx)$
$=\frac{\sin(2^kx)}{2^k\sin(x)}\times\cos(2^kx)$ (using Step 2)
$=\frac{\sin(2^kx)\cos(2^kx)}{2^k\sin(x)}$
$=\frac{2\sin(2^kx)\cos(2^kx)}{2^{k+1}\sin(x)}$
$=\frac{\sin(2\times2^kx)}{2^{k+1}\sin(x)}$
$=\frac{\sin(2^{k+1}x)}{2^{k+1}\sin(x)}$
$=RHS$
Step 4 - Conclusion
If true for $n=k$, then we have proven it is true for $n=k+1$.
Also, as true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
b) Answer and Solution are the same for proofs
To prove: $\cos\left(\frac{x}{2}\right) + \cos\left(\frac{3x}{2}\right)+\cdots + \cos\left(\frac{(2n-1)x}{2}\right) = \frac{\sin nx}{2\sin\left(\frac x 2\right)}$ for $n\ge1$ and $n\in\mathbb{Z}^+$:
Step 1 - Prove true for $n=1$
$LHS=\cos\left(\frac x2\right)$
$RHS=\frac{\sin x}{2\sin\left(\frac x 2\right)}$
$=\frac{2\sin\left(\frac x2\right)\cos\left(\frac x2\right)}{2\sin\left(\frac x 2\right)}$
$=\cos\left(\frac x2\right)$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\cos\left(\frac{x}{2}\right) + \cos\left(\frac{3x}{2}\right)+\cdots + \cos\left(\frac{(2k-1)x}{2}\right) = \frac{\sin kx}{2\sin\left(\frac x 2\right)}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\cos\left(\frac{x}{2}\right) + \cos\left(\frac{3x}{2}\right)+\cdots + \cos\left(\frac{(2k-1)x}{2}\right) + \cos\left(\frac{(2(k+1)-1)x}{2}\right) = \frac{\sin(k+1)x}{2\sin\left(\frac x 2\right)}$
$LHS=\cos\left(\frac{x}{2}\right) + \cos\left(\frac{3x}{2}\right)+\cdots + \cos\left(\frac{(2k-1)x}{2}\right) + \cos\left(\frac{(2k+1)x}{2}\right)$
$=\frac{\sin kx}{2\sin\left(\frac x 2\right)}+ \cos\left(\frac{(2k+1)x}{2}\right)$ (using Step 2)
$=\frac{\sin kx + 2\sin\left(\frac x 2\right)\cos\left(kx+\frac x 2\right)}{2\sin\left(\frac x 2\right)}$
$=\frac{\sin kx + 2\sin\left(\frac x 2\right)\left[\cos kx\cos\left(\frac x 2\right)-\sin kx\sin\left(\frac x 2\right)\right]}{2\sin\left(\frac x 2\right)}$
$\frac{\sin kx \left[1-2\sin^2\left(\frac x 2\right)\right]+2\sin\left(\frac x 2\right)\cos\left(\frac x 2\right)\cos kx}{2\sin\left(\frac x 2\right)}$
$=\frac{\sin kx\cos x+\sin x\cos kx}{2\sin\left(\frac x 2\right)}$
$=\frac{\sin(kx+x)}{2\sin\left(\frac x 2\right)}$
$=\frac{\sin(k+1)x}{2\sin\left(\frac x 2\right)}$
$=RHS$
Step 4 - Conclusion
If true for $n=k$, then we have proven it is true for $n=k+1$.
Also, as true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
c) Answer and Solution are the same for proofs
To prove: $\frac12+\cos x+\cos 2x+\dotsb+\cos nx=\frac{\sin\left(n+\frac12\right)x}{2\sin\frac12x}$
Step 1 - Prove true for $n=1$
$LHS=\frac12+\cos x$
$RHS=\frac{\sin\left(1+\frac12\right)x}{2\sin\frac12x}$
$=\frac{\sin\frac32x}{2\sin\frac12x}$
$=\frac{3\sin\frac12x-4\sin^3\frac12x}{2\sin\frac12x}$
$=\frac{3-4\sin^2\frac12x}{2}$
$=\frac{3-2(1-\cos x)}{2}$
$=\frac{1+2\cos x}{2}$
$=\frac12+\cos x$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\frac12+\cos x+\cos 2x+\dotsb+\cos kx=\frac{\sin\left(k+\frac12\right)x}{2\sin\frac12x}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\frac12+\cos x+\cos 2x+\dotsb+\cos kx+\cos (k+1)x=\frac{\sin\left(k+1+\frac12\right)x}{2\sin\frac12x}$
$LHS=\frac12+\cos x+\cos 2x+\dotsb+\cos kx+\cos (k+1)x$
$=[\frac12+\cos x+\cos 2x+\dotsb+\cos kx]\,+\cos (k+1)x$
$=[\frac12+\cos x+\cos 2x+\dotsb+\cos kx]\,+\cos (k+1)x$
$=\frac{\sin\left(k+\frac12\right)x}{2\sin\frac12x}+\cos (k+1)x$ (using Step 2)
$=\frac{\sin\left(k+\frac12\right)x}{2\sin\frac12x}+\frac{2\cos (k+1)x\sin\frac12x}{2\sin\frac12x}$
$=\frac{\sin\left(k+\frac12\right)x+2\cos (k+1)x\sin\frac12x}{2\sin\frac12x}$
$=\frac{\sin\left(k+\frac12\right)x+\sin\left(k+\frac32\right)x-\sin\left(k-\frac12\right)x}{2\sin\frac12x}$
$=\frac{\sin\left(k+1+\frac12\right)x}{2\sin\frac12x}$
$=RHS$
Step 4 - Conclusion
If true for $n=k$, then we have proven it is true for $n=k+1$.
Also, as true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
d) Answer and Solution are the same for proofs
To prove: $\sin x+\sin 2x+\sin 3x+\dots+\sin nx=\frac{\sin\frac12(n+1)x\sin\frac12nx}{\sin\frac12x}$ for all positive integers, $n$
Step 1 - Prove true for $n=1$
$LHS=\sin x$
$RHS=\frac{\sin\frac12\times2\times x\times\sin\frac12\times1\times x}{\sin\frac12x}$
$=\frac{\sin x\sin\frac12x}{\sin\frac12x}$
$=\sin x$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\sin x+\sin 2x+\sin 3x+\dots+\sin kx=\frac{\sin\frac12(k+1)x\sin\frac12kx}{\sin\frac12x}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\sin x+\sin 2x+\sin 3x+\dots+\sin kx+\sin(k+1)x=\frac{\sin\frac12(k+2)x\sin\frac12(k+1)x}{\sin\frac12x}$
$LHS=\sin x+\sin 2x+\sin 3x+\dots+\sin kx+\sin(k+1)x$
$=\frac{\sin\frac12(k+1)x\sin\frac12kx}{\sin\frac12x}+\sin(k+1)x$ (using Step 2)
$=\frac{\sin\frac12(k+1)x\sin\frac12kx+\sin(k+1)x\sin\frac12x}{\sin\frac12x}$
Aside:
Use $\sin A\sin B=\frac12\left(\cos(A-B)-\cos(A+B)\right)$
so $=\frac{1}{\sin\frac12 x}\times\left[\frac12\left(\cos\frac12x-\cos(k+\frac12)x\right)+\frac12\left(\cos(k+\frac12)x-\cos(k+\frac32)x\right)\right]$
$=\frac{1}{\sin\frac12 x}\times\frac12\left(\cos\frac12x-\cos(k+\frac32)x\right)$
$=\frac{1}{\sin\frac12 x}\times\sin\frac{(k+1)x}{2}\sin\frac{(k+2)x}{2}$
$=RHS$
Step 4 - Conclusion
If true for $n=k$, then we have proven it is true for $n=k+1$.
Also, as true when $n=1$, it is also true for all positive integers by mathematical induction.
Question 2
a) Answer and Solution are the same for proofs
To prove: $(\cos(x)+i\sin(x))^n=\cos(nx)+i\sin(nx)$ for $n\ge0$ and $n\in\mathbb{Z}^+$:
Step 1 - Prove true for $n=0$
$LHS=(\cos(x)+i\sin(x))^0$
$=1$
$RHS=\cos(0)+i\sin(0)$
$=1+0$
$=1$
$=LHS$
So true for $n=0$
Step 2 - Assume true for $n=k$
$(\cos(x)+i\sin(x))^k=\cos(kx)+i\sin(kx)$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$(\cos(x)+i\sin(x))^{k+1}=\cos((k+1)x)+i\sin((k+1)x)$
$LHS=(\cos(x)+i\sin(x))^{k+1}$
$=(\cos(x)+i\sin(x))^k((\cos(x)+i\sin(x))$
$=(\cos(kx)+i\sin(kx))((\cos(x)+i\sin(x))$ (using Step 2)
$=\cos(kx)\cos(x)+i\cos(kx)\sin(x)+i\cos(x)\sin(kx)-\sin(kx)\sin(x)$
$=\cos(kx)\cos(x)-\sin(kx)\sin(x)+i(\cos(kx)\sin(x)+\cos(x)\sin(kx))$
$=\cos(kx+x)+i\sin(kx+x)$
$=\cos((k+1)x)+i\sin((k+1)x)$
$=RHS$
Step 4 - Conclusion
If true for $n=k$, then we have proven it is true for $n=k+1$.
Also, as true when $n=0$, it is also true for all $n\ge0$ and $n\in\mathbb{Z}^+$ by mathematical induction.
Solutions
Question 1
a) Answer and Solution are the same for proofs
To prove: $\cos x\times\cos 2x\times\dots\times\cos(2^{n-1}x)=\frac{\sin(2^nx)}{2^n\sin(x)}$ for $n\ge1$ and $n\in\mathbb{Z}^+$:
Step 1 - Prove true for $n=1$
$LHS=\cos x$
$RHS=\frac{\sin(2x)}{2\sin x}$
$=\frac{2\sin x\cos x}{2\sin x}$
$=\cos x$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\cos x\times\cos 2x\times\dots\times\cos(2^{k-1}x)=\frac{\sin(2^kx)}{2^k\sin(x)}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\cos x\times\cos 2x\times\dots\times\cos(2^{(k+1)-1}x)=\frac{\sin(2^{k+1}x)}{2^{k+1}\sin(x)}$
$LHS=\cos x\times\cos 2x\times\dots\times\cos(2^{k-1}x)\times\cos(2^kx)$
$=\frac{\sin(2^kx)}{2^k\sin(x)}\times\cos(2^kx)$ (using Step 2)
$=\frac{\sin(2^kx)\cos(2^kx)}{2^k\sin(x)}$
$=\frac{2\sin(2^kx)\cos(2^kx)}{2^{k+1}\sin(x)}$
$=\frac{\sin(2\times2^kx)}{2^{k+1}\sin(x)}$
$=\frac{\sin(2^{k+1}x)}{2^{k+1}\sin(x)}$
$=RHS$
Step 4 - Conclusion
If true for $n=k$, then we have proven it is true for $n=k+1$.
Also, as true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
b) Answer and Solution are the same for proofs
To prove: $\cos\left(\frac{x}{2}\right) + \cos\left(\frac{3x}{2}\right)+\cdots + \cos\left(\frac{(2n-1)x}{2}\right) = \frac{\sin nx}{2\sin\left(\frac x 2\right)}$ for $n\ge1$ and $n\in\mathbb{Z}^+$:
Step 1 - Prove true for $n=1$
$LHS=\cos\left(\frac x2\right)$
$RHS=\frac{\sin x}{2\sin\left(\frac x 2\right)}$
$=\frac{2\sin\left(\frac x2\right)\cos\left(\frac x2\right)}{2\sin\left(\frac x 2\right)}$
$=\cos\left(\frac x2\right)$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\cos\left(\frac{x}{2}\right) + \cos\left(\frac{3x}{2}\right)+\cdots + \cos\left(\frac{(2k-1)x}{2}\right) = \frac{\sin kx}{2\sin\left(\frac x 2\right)}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\cos\left(\frac{x}{2}\right) + \cos\left(\frac{3x}{2}\right)+\cdots + \cos\left(\frac{(2k-1)x}{2}\right) + \cos\left(\frac{(2(k+1)-1)x}{2}\right) = \frac{\sin(k+1)x}{2\sin\left(\frac x 2\right)}$
$LHS=\cos\left(\frac{x}{2}\right) + \cos\left(\frac{3x}{2}\right)+\cdots + \cos\left(\frac{(2k-1)x}{2}\right) + \cos\left(\frac{(2k+1)x}{2}\right)$
$=\frac{\sin kx}{2\sin\left(\frac x 2\right)}+ \cos\left(\frac{(2k+1)x}{2}\right)$ (using Step 2)
$=\frac{\sin kx + 2\sin\left(\frac x 2\right)\cos\left(kx+\frac x 2\right)}{2\sin\left(\frac x 2\right)}$
$=\frac{\sin kx + 2\sin\left(\frac x 2\right)\left[\cos kx\cos\left(\frac x 2\right)-\sin kx\sin\left(\frac x 2\right)\right]}{2\sin\left(\frac x 2\right)}$
$\frac{\sin kx \left[1-2\sin^2\left(\frac x 2\right)\right]+2\sin\left(\frac x 2\right)\cos\left(\frac x 2\right)\cos kx}{2\sin\left(\frac x 2\right)}$
$=\frac{\sin kx\cos x+\sin x\cos kx}{2\sin\left(\frac x 2\right)}$
$=\frac{\sin(kx+x)}{2\sin\left(\frac x 2\right)}$
$=\frac{\sin(k+1)x}{2\sin\left(\frac x 2\right)}$
$=RHS$
Step 4 - Conclusion
If true for $n=k$, then we have proven it is true for $n=k+1$.
Also, as true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
c) Answer and Solution are the same for proofs
To prove: $\frac12+\cos x+\cos 2x+\dotsb+\cos nx=\frac{\sin\left(n+\frac12\right)x}{2\sin\frac12x}$
Step 1 - Prove true for $n=1$
$LHS=\frac12+\cos x$
$RHS=\frac{\sin\left(1+\frac12\right)x}{2\sin\frac12x}$
$=\frac{\sin\frac32x}{2\sin\frac12x}$
$=\frac{3\sin\frac12x-4\sin^3\frac12x}{2\sin\frac12x}$
$=\frac{3-4\sin^2\frac12x}{2}$
$=\frac{3-2(1-\cos x)}{2}$
$=\frac{1+2\cos x}{2}$
$=\frac12+\cos x$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\frac12+\cos x+\cos 2x+\dotsb+\cos kx=\frac{\sin\left(k+\frac12\right)x}{2\sin\frac12x}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\frac12+\cos x+\cos 2x+\dotsb+\cos kx+\cos (k+1)x=\frac{\sin\left(k+1+\frac12\right)x}{2\sin\frac12x}$
$LHS=\frac12+\cos x+\cos 2x+\dotsb+\cos kx+\cos (k+1)x$
$=[\frac12+\cos x+\cos 2x+\dotsb+\cos kx]\,+\cos (k+1)x$
$=[\frac12+\cos x+\cos 2x+\dotsb+\cos kx]\,+\cos (k+1)x$
$=\frac{\sin\left(k+\frac12\right)x}{2\sin\frac12x}+\cos (k+1)x$ (using Step 2)
$=\frac{\sin\left(k+\frac12\right)x}{2\sin\frac12x}+\frac{2\cos (k+1)x\sin\frac12x}{2\sin\frac12x}$
$=\frac{\sin\left(k+\frac12\right)x+2\cos (k+1)x\sin\frac12x}{2\sin\frac12x}$
$=\frac{\sin\left(k+\frac12\right)x+\sin\left(k+\frac32\right)x-\sin\left(k-\frac12\right)x}{2\sin\frac12x}$
$=\frac{\sin\left(k+1+\frac12\right)x}{2\sin\frac12x}$
$=RHS$
Step 4 - Conclusion
If true for $n=k$, then we have proven it is true for $n=k+1$.
Also, as true when $n=1$, it is also true for all $n\ge1$ and $n\in\mathbb{Z}^+$ by mathematical induction.
d) Answer and Solution are the same for proofs
To prove: $\sin x+\sin 2x+\sin 3x+\dots+\sin nx=\frac{\sin\frac12(n+1)x\sin\frac12nx}{\sin\frac12x}$ for all positive integers, $n$
Step 1 - Prove true for $n=1$
$LHS=\sin x$
$RHS=\frac{\sin\frac12\times2\times x\times\sin\frac12\times1\times x}{\sin\frac12x}$
$=\frac{\sin x\sin\frac12x}{\sin\frac12x}$
$=\sin x$
$=LHS$
So true for $n=1$
Step 2 - Assume true for $n=k$
$\sin x+\sin 2x+\sin 3x+\dots+\sin kx=\frac{\sin\frac12(k+1)x\sin\frac12kx}{\sin\frac12x}$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$\sin x+\sin 2x+\sin 3x+\dots+\sin kx+\sin(k+1)x=\frac{\sin\frac12(k+2)x\sin\frac12(k+1)x}{\sin\frac12x}$
$LHS=\sin x+\sin 2x+\sin 3x+\dots+\sin kx+\sin(k+1)x$
$=\frac{\sin\frac12(k+1)x\sin\frac12kx}{\sin\frac12x}+\sin(k+1)x$ (using Step 2)
$=\frac{\sin\frac12(k+1)x\sin\frac12kx+\sin(k+1)x\sin\frac12x}{\sin\frac12x}$
Aside:
Use $\sin A\sin B=\frac12\left(\cos(A-B)-\cos(A+B)\right)$
so $=\frac{1}{\sin\frac12 x}\times\left[\frac12\left(\cos\frac12x-\cos(k+\frac12)x\right)+\frac12\left(\cos(k+\frac12)x-\cos(k+\frac32)x\right)\right]$
$=\frac{1}{\sin\frac12 x}\times\frac12\left(\cos\frac12x-\cos(k+\frac32)x\right)$
$=\frac{1}{\sin\frac12 x}\times\sin\frac{(k+1)x}{2}\sin\frac{(k+2)x}{2}$
$=RHS$
Step 4 - Conclusion
If true for $n=k$, then we have proven it is true for $n=k+1$.
Also, as true when $n=1$, it is also true for all positive integers by mathematical induction.
Question 2
a) Answer and Solution are the same for proofs
To prove: $(\cos(x)+i\sin(x))^n=\cos(nx)+i\sin(nx)$ for $n\ge0$ and $n\in\mathbb{Z}^+$:
Step 1 - Prove true for $n=0$
$LHS=(\cos(x)+i\sin(x))^0$
$=1$
$RHS=\cos(0)+i\sin(0)$
$=1+0$
$=1$
$=LHS$
So true for $n=0$
Step 2 - Assume true for $n=k$
$(\cos(x)+i\sin(x))^k=\cos(kx)+i\sin(kx)$
Step 3 - Show true for $n=k+1$
If true for $n=k+1$ then need to show:
$(\cos(x)+i\sin(x))^{k+1}=\cos((k+1)x)+i\sin((k+1)x)$
$LHS=(\cos(x)+i\sin(x))^{k+1}$
$=(\cos(x)+i\sin(x))^k((\cos(x)+i\sin(x))$
$=(\cos(kx)+i\sin(kx))((\cos(x)+i\sin(x))$ (using Step 2)
$=\cos(kx)\cos(x)+i\cos(kx)\sin(x)+i\cos(x)\sin(kx)-\sin(kx)\sin(x)$
$=\cos(kx)\cos(x)-\sin(kx)\sin(x)+i(\cos(kx)\sin(x)+\cos(x)\sin(kx))$
$=\cos(kx+x)+i\sin(kx+x)$
$=\cos((k+1)x)+i\sin((k+1)x)$
$=RHS$
Step 4 - Conclusion
If true for $n=k$, then we have proven it is true for $n=k+1$.
Also, as true when $n=0$, it is also true for all $n\ge0$ and $n\in\mathbb{Z}^+$ by mathematical induction.