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there is this question in my assignment that i'm struggling to do, it says:

Using Euler's formula, prove that: $$\cos(2\theta)+\cos(4\theta)+\cdots+\cos(2n\theta)=\frac{\sin(n\theta)}{\sin(\theta)}\cos((n+1)\theta)$$

I was unable to get anywhere with expanding the LHS, so i have attempted to arrive at the LHS starting from the RHS, and this is what i have got to so far: $$\sin(\theta)=\frac{e^{i\theta}-e^{-i\theta}}{2i}$$ $$\sin(n\theta)=\frac{e^{in\theta}-e^{-in\theta}}{2i}$$ $$\cos((n+1)\theta)=\frac{e^{i(n+1)\theta}+e^{-i(n+1)\theta}}{2}$$ and by multiplying the last two, then dividing by the second one, i arrived at: $$\frac{1}{2}(\frac{e^{i(2n+1)\theta}-e^{-i(2n+1)\theta}}{e^{i\theta}-e^{-i\theta}}-1)$$ and i am struggling to move any further than this, i assume there's an easier way to do this by starting from the LHS instead, but i couldn't figure it out.

Sebastiano
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Momo
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3 Answers3

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Euler's formula:

$$e^{ix}=\cos x + i\sin x.$$

This implies $$e^{inx}= \cos(nx) +i \sin(nx)= (\cos x+ i \sin x)^n.$$

Observe that $$ x^2+x^4+\dots+x^{2n} = x^2 \frac{x^{2n}-1}{x^2-1}.$$

Set $x=e^{i\theta}$. We would then easily see that $$\Re(LHS)=\cos(2\theta)+\cos(4\theta)+\dots+\cos(2n\theta), $$ where $\Re$ stands for real part. Also see that

$$RHS = \frac{x^{2n}-1}{1-x^{-2}}=\left(\frac{\cos(2n\theta)-1+i\sin(2n\theta)}{1-(\cos(2\theta)-i\sin(2\theta))}\right),$$ as $x^{-2}=e^{-2i\theta} = \cos(2\theta)-i\sin(2\theta)$.

Reduce the right hand side to $a+ib$ form by multiplying and dividing by the conjugate of the denominator and using the formulas $1-\cos2x= 2\sin^2x$ and formula of $\cos(A+B)$. You would then observe that $$a=\Re(RHS) = \frac{\sin(n\theta)\cos((n+1)\theta)}{\sin(\theta)}.$$

Yathi
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It is rather easy to prove the identity using Product-to-sum formula.

Multiplying the sum by $2\sin \theta$ gives a telescoping series as $$ \begin{aligned} &2 \sin \theta[\cos (2 \theta)+\cos (4 \theta)+\ldots+\cos (2 n \theta)] \\ = & {[\sin (3 \theta)-\sin \theta]+[\sin (5 \theta)-\sin (3 \theta)]+\cdots+ [\sin (2 n+1) \theta-\sin (2 n-1) \theta] } \\ = & \sin (2 n+1) \theta-\sin \theta \\ = & 2 \cos (n+1) \theta \sin (n \theta) \end{aligned} $$ Re-arranging gives $$ \cos(2 \theta)+\cos (4 \theta)+\cdots+\cos (2 n \theta)=\frac{\sin (n \theta) \cos (n+1)}{\sin \theta} $$

Lai
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By Euler formula $$ \begin{aligned} \sum_{k=1}^n \cos (2 k \theta)&=\sum_{k=1}^n \frac{e^{2 k \theta i}+e^{-2 k \theta i}}{2} \\ & =\frac{1}{2}\left[\sum_{k=1}^n e^{2 k \theta i}+\sum_{k=1}^n e^{-2 k \theta i}\right] \\ & =\frac{1}{2}\left[\frac{e^{2 \theta i}\left(e^{2 n \theta i}-1\right)}{e^{2 \theta i}-1}+\frac{e^{-2 \theta i}\left(e^{-2 n \theta i}-1\right)}{e^{-2 \theta i}-1}\right] \\ &= \frac{1}{2}\left[\frac{e^{(2 n+1) \theta i}-e^{\theta i}}{e^{\theta i}-e^{-\theta i}}-\frac{e^{-(2 n+1) \theta i}-e^{-\theta i}}{e^{\theta i}-e^{-\theta i}}\right]\\&= \frac{1}{2\left(e^{\theta i}-e^{-\theta i}\right)}\left[(e^{(2 n+1) \theta i}-e^{-(2 n+1) \theta i})-\left(e^{\theta i}-e^{-\theta i}\right)\right]\\&=\frac{1}{4 i \sin \theta}[2 i \sin (2 n+1) \theta-2 i \sin \theta]\\&= \frac{cos(n+1) \theta \sin (n \theta)}{\sin \theta} \end{aligned} $$

Lai
  • 20,421