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Prove that:$$\cos x+\cos(2x)+\cdots+\cos (nx)=\frac{\sin(\frac{nx}{2})\cos\frac{(n+1)x}{2}}{\sin(\frac{x}{2})}.\ (1)$$

My attempt:$$\sin\left(\frac{x}{2}\right)\sum_{k=1}^{n}\cos{(kx)}$$$$=\sum_{k=1}^{n}\sin\left(\frac{x}{2}\right)\cos{(kx)} $$ (Applying $\sin x\cos y =\frac{1}{2}\left[\sin(x+y)+\sin(x-y)\right]$) $$=\sum_{k=1}^{n}\frac{1}{2}\sin\left(\frac{x}{2}-kx\right)+\frac{1}{2}\sin\left(\frac{x}{2}+kx\right).$$Multiplying $(1)$ by $\sin\left(\frac{x}{2}\right)$ ,gives: $$\sum_{k=1}^{n}\sin\left(\frac{x}{2}\right)\cos(kx)=\sin\left(\frac{nx}{2}\right)\cos{\frac{(n+1)x}{2}}$$ $$\Leftrightarrow\sum_{k=1}^{n}\left[\sin\left(\frac{x}{2}-kx\right)+\sin\left(\frac{x}{2}+kx\right)\right]=\sin\left(-\frac{x}{2}\right)+\sin\left(\frac{x}{2}+nx\right).$$Then, by induction on $n$ and using $(\sin x\sin y)$ and $(\sin x +\sin y)$ formulas, I end in here: $$\sum_{k=1}^{n+1}\sin\left(\frac{x}{2}\right)\cos(kx)=\left[2\sin\left(\frac{nx}{2}\right)\cos\frac{(n+1)x}{2}\right]+\left[2\sin\left(\frac{x}{2}\right)\cos(n+1)x\right].$$ Any help would be appreciated, thanks!

Blue
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user180321
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2 Answers2

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Here is the induction step: it comes down to proving \begin{gather*}\frac{\sin \dfrac{nx}{2} \cos\dfrac{(n+1)x}{2}}{\sin\dfrac{x}{2}}+\cos(n+1)x=\frac{\sin\dfrac{(n+1)x}{2}\cos\dfrac{(n+2)x}{2}}{\sin(\dfrac{x}{2})}\\ \text{or}\qquad\sin\dfrac{nx}{2}\cos\dfrac{(n+1)x}{2}+\sin\dfrac{x}{2}\cos(n+1)x=\sin\dfrac{(n+1)x}{2} \cos\dfrac{(n+2)x}{2} \end{gather*} Now use the linearisation formulae: \begin{cases}\displaystyle \sin\frac{nx}{2}\cos\frac{(n+1)x}{2}=\tfrac12\biggl(\sin\frac{(2n+1)x}{2}-\sin\frac x2\biggr),\\[1ex] \sin\dfrac{x}{2}\cos(n+1)x=\frac12\biggl(\sin\dfrac{(2n+3)x}{2}-\sin\dfrac{(2n+1)x}{2}\biggr), \end{cases} whence the l.h.s. is $$\frac12\biggl(\sin\dfrac{(2n+3)x}{2}-\sin\frac x2\biggr)=\sin\frac{(n+1)x}2\,\cos\frac{(n+2)x}2$$ by the factorisation formula: $\;\sin p-\sin q=2\sin\dfrac{p-q}2\cos\dfrac{p+q}2$.

Bernard
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  • Since when $$\sin\frac{nx}{2}\cos\frac{(n+1)x}{2}=\frac{1}{2}\left(\sin\frac{(2n+1)x}{2}- \sin\frac{x}{2}\right)$$ and $$\sin\frac{x}{2}\cos(n+1)x=\frac{1}{2}\left(\sin\frac{(2n+3)x}{2}-\sin\frac{(2n+1)x}{2}\right)$$ ? – user180321 Apr 19 '16 at 21:44
  • Apply the linearisation formula: $; \sin a \cos b=\frac12\bigl(\sin(a+b)+\sin(a-b)\bigr)$. – Bernard Apr 19 '16 at 21:48
  • Well, it was easier than I thought – user180321 Apr 19 '16 at 21:53
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    It's easier with Euler's formulae and complex numbers, though. With trigonometry you have to know a little more than basic formulae. Actually I think the minimal background in trigonometry is ~ 20 formulae ( not counting this one, which everyone should know). – Bernard Apr 19 '16 at 21:58
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You want to show that $$ \frac{\sin\frac{nx}{2}\cos\frac{(n+1)x}{2}}{\sin\frac{x}{2}} +\cos((n+1)x)= \frac{\sin\frac{(n+1)x}{2}\cos\frac{(n+2)x}{2}}{\sin\frac{x}{2}} $$ which is the same as $$ \sin\frac{nx}{2}\cos\frac{(n+1)x}{2}+\sin\frac{x}{2}\cos((n+1)x)= \sin\frac{(n+1)x}{2}\cos\frac{(n+2)x}{2} $$ Set $s=nx/2$ and $t=x/2$; the relation becomes, after multiplying by $2$ for later applying the product to sum formulas, $$ 2\sin s\cos(s+t)+2\sin t\cos(2s+2t)=2\sin(s+t)\cos(s+2t) $$ Recall $$ 2\sin\alpha\cos\beta=\sin(\alpha+\beta)+\sin(\alpha-\beta) $$ so the left-hand side becomes $$ \sin(2s+t)+\sin(-t)+\sin(2s+3t)+\sin(-2s-t)= \sin(2s+3t)-\sin t $$ The right-hand side becomes $$ \sin(2s+3t)+\sin(-t)=\sin(2s+3t)-\sin t $$

egreg
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