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Lately a friend of mine asked me above question.

After reaching this

$$-\cos(12)+\cos(24)-\cos(36)+\cos(48)-\cos(60)+\cos(72)-\cos(84)+ 1, $$

I could not simply it any further but later I noticed that there is a general pattern here
$L=\sin^2(x)- \sin^2(2x)+ \sin^2 (3x)-\cdots\pm\sin^2 (kx)$,

where $kx=90^\circ$ and $k \in \mathbb{R} $

$L = \cfrac{1}{2} \text{ for $k$ odd}$

$L= -\cfrac{1}{2} \text{ for $k$ even}$

And $M=\cos^2(x)- \cos^2(2x)+ \cos^2 (3x) - \cdots \pm \cos^2 (kx)$

$M= \dfrac{1}{2}$ for $k=1,2,3,\ldots$

Can anyone explain it to me?

Michael Hardy
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ravi
  • 11

3 Answers3

3

We have that $$\sum_{k=1}^{15}(-1)^k\sin^2(6°k)=\frac{1}{2}\sum_{k=1}^{15}(-1)^k(1-\cos(12°k))=-\frac{1}{2}+S=-\frac{1}{2}$$ where $S=0$ because $$S=\sum_{k=1}^{15}(-1)^k \cos(12°k)=\mbox{Re}\left(\sum_{k=1}^{15}(-1)^ke^{ik\pi/15}\right)\\=\mbox{Re}\left(-e^{i\pi/15}\cdot \frac{1+e^{i15\pi/15}}{1+e^{i\pi/15}}\right)=\mbox{Re}(0)=0.$$

Robert Z
  • 145,942
1

Hint: You can rewrite the first equation because ${sin}^2(kx)=\dfrac{1}{2}(1-cos(2kx))$: $$\sum_{k=0}^n{(-1)^n\sin^2(kx)}=\sum_{k=0}^n(-1)^n{\dfrac{1}{2}}(1-cos(2kx))$$ $$\sum_{k=0}^n(-1)^n{\dfrac{1}{2}}(1-cos(2kx))=\sum_{k=0}^n(-1)^n{\dfrac{1}{2}}-\dfrac{1}{2}\sum_{k=0}^n(-1)^ncos(2kx)$$

Curious
  • 864
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If $S=\cos12^\circ-\cos24^\circ+\cos36^\circ-\cdots$ upto $n$ terms

As $\cos(180^\circ+y)=-\cos y,\cos(360^\circ+y)=+\cos y$

$$-S=\sum_{r=1}^n\cos\{r(180+12)^\circ\}$$

Using $\sum \cos$ when angles are in arithmetic progression,

$$-S=\dfrac{\sin\left(n\cdot\dfrac{192^\circ}2\right)\cos\left(192^\circ+\dfrac{(n-1)192^\circ}2\right)}{\sin\dfrac{192^\circ}2}=\dfrac{\sin(n\cdot96^\circ)\cos(n+1)96^\circ}{\sin96^\circ}$$

Here $n=15$

Community
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lab bhattacharjee
  • 274,582