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Prove that $\cos{2\alpha}+\cos{4\alpha}+\cdots+\cos{2n\alpha} = \dfrac{\sin{(n+1)\alpha}\cos{n\alpha}}{\sin{\alpha}}-1$.

What is wrong with this proof?

Let $S = \cos{2\alpha}+\cos{4\alpha}+\cdots+\cos{2n\alpha}$. Then, $$S\sin{\alpha} = \sin{\alpha}\cos{2\alpha}+\sin{\alpha}\cos{4\alpha}+\cdots+\sin{\alpha}\cos{2n\alpha}$$

$$S\sin{\alpha} = \dfrac{1}{2}\left[\left (\sin{\dfrac{3\alpha}{2}-\sin{\dfrac{1}{2}\alpha}}\right )+\left(\sin{\dfrac{5\alpha}{2}}-\sin{\dfrac{3}{2}\alpha} \right)+\cdots\right] = \dfrac{\sin{\dfrac{\alpha(n+1)}{2}-\sin \left (\dfrac{1}{2}\alpha \right)}}{2}.$$

Thus, $$S = \dfrac{\sin{\dfrac{\alpha(n+1)}{2}-\sin \left (\dfrac{1}{2}\alpha \right)}}{2\sin{\alpha}}.$$

According to my answer key, the answer is $$\dfrac{\sin{(n+1)\alpha \cos{n\alpha}}}{\sin{\alpha}}-1.$$

Did I make an error somewhere?

Edit: I made a mistake with the $\sin(x)-\sin(y)$ formula above. I should say the following: $$S\sin{\alpha} = \sin{\alpha}\cos{2\alpha}+\sin{\alpha}\cos{4\alpha}+\cdots+\sin{\alpha}\cos{2n\alpha} = \dfrac{1}{2}\left[\left (\sin{3\alpha-\sin{\alpha}}\right )+\left(\sin{5\alpha}-\sin{3\alpha} \right)+\cdots\right] = \dfrac{\sin{(2n+1)-\sin(\alpha)}}{2}.$$

Thus, $S = \dfrac{\sin{(2n+1)\alpha-\sin(\alpha)}}{2\sin{\alpha}}.$

user19405892
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  • The formula you try to apply reads (for the first term) $\sin(\alpha)\cos(2\alpha) = \frac{1}{2}[\sin \left(3\alpha\right)-\sin\left(\alpha\right)]$. You have $3\alpha/2$ and $\alpha/2$ instead on the right hand side. – Winther Jan 10 '16 at 04:06
  • @Winther There is no times 2 though. – user19405892 Jan 10 '16 at 04:08
  • Hint:

    $$\sin x\cos y=\frac{1}{2}\left(\sin(x+y)+\sin(x-y)\right)=\frac{1}{2}\sin(x+y)-\frac{1}{2} \sin(y-x)$$ Implies \begin{align} S\sin \alpha&=\sum_{k=1}^n\left[\frac{1}{2}\sin\left(2k+1\right)\alpha-\frac{1}{2} \sin\left(2k-1\right)\alpha\right] \end{align} Which is telescopic.

     – Ángel Mario Gallegos Jan 10 '16 at 04:11
  • @Winther Ah, I see what you mean. In that case carrying out the computations I should get the right result? – user19405892 Jan 10 '16 at 04:12
  • @Winther Check my edit and tell me if that's right. – user19405892 Jan 10 '16 at 04:25
  • The formula you have found is correct, however you still need to show that $\frac{\sin[(2n+1)x]-\sin(x)}{2\sin(x)} = \frac{\sin[(n+1)x]\cos(nx)-\sin(x)}{\sin(x)}$ to solve the problem you started with. – Winther Jan 10 '16 at 04:27
  • My attempt: since $\sin[(2n+1)x]-\sin(x) = 2\sin(n)\cos[(n+1)x]$, we have $\dfrac{\sin[(2n+1)x]-\sin(x)}{2\sin(x)} = \dfrac{2\sin(n)\cos[(n+1)x]}{2\sin(x)} = \dfrac{\sin(n)\cos[(n+1)x]}{\sin(x)}$. Then what? – user19405892 Jan 10 '16 at 04:31
  • First notice that you can write this as $\sin[(n+1)x]\cos(nx) = \frac{1}{2}[\sin[(2n+1)x] + \sin(x)]$. Then try to apply some known identity for the product of $\sin(x)\cos(y)$ with $x,y$ choosen appropriately. The needed identity can be derived from the sum-formulas for $\sin$ and $\cos$. – Winther Jan 10 '16 at 04:34
  • Where are you getting $\sin[(n+1)x]$ from? – user19405892 Jan 10 '16 at 04:35
  • Wolfram seems to disagree with us: http://tinyurl.com/jnzycgc – user19405892 Jan 10 '16 at 04:37
  • It's true for all $x\in\mathbb{R}$ and $n\in\mathbb{N}$ which is what you are looking for. Do the derivation and you will see. Don't fixate too much about what Wolfram is saying. – Winther Jan 10 '16 at 04:41
  • Ah, I see again. Thus, to finish it we have $\frac{\sin[(n+1)x]\cos(nx)-\sin(x)}{\sin(x)} = \frac{\frac{1}{2}[\sin[(2n+1)x]+\sin(x)]}{\sin(x)}$. What is the last step? – user19405892 Jan 10 '16 at 04:52
  • See comment above "First notice that... Then try to...". Also the formula above has a typo: the $+\sin(x)$ on the rhs should be $-\sin(x)$. – Winther Jan 10 '16 at 04:54
  • Got it! Great thanks. – user19405892 Jan 10 '16 at 04:59

1 Answers1

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Now that you've gotten $S\sin \alpha = \dfrac{1}{2}\left[\sin(2n+1)\alpha-\sin\alpha\right]$, use the difference to product identity $\sin A - \sin B = 2\sin\dfrac{A+B}{2}\cos\dfrac{A+B}{2}$ to get $S\sin \alpha = \sin n\alpha \cos (n+1)\alpha$.

This isn't exactly what's on the right side of the given inequality. But you can use the sum of angles identity to get $\sin \alpha = \sin(n+1)\alpha\cos n\alpha - \sin n\alpha\cos(n+1)\alpha$.

Can you finish it from here?

JimmyK4542
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