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10 degrees

$$\sin^2(10^\circ)-\sin^2(20^\circ)-\sin^2(40^\circ)=-\frac{1}{2}$$

$$\cos^2(10^\circ)-\cos^2(20^\circ)-\cos^2(40^\circ)=-\frac{1}{2}$$

Why are they both have same answer?

The only time they have same answer is at 45 degrees right?

$\sin(45^\circ)=\cos(45^\circ)=\frac{1}{\sqrt2}$

Can somebody provide me an explanation please?

Also how to prove these two identities

I know all others $15^\circ, 30^\circ, 45^\circ, 60^\circ$, etc, but can't seem to prove these.

Zhanxiong
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    Same answer because $\sin^2x+\cos^2x=1$, so the sum of the two right hand sides has to be $-1$. – almagest May 19 '16 at 04:35
  • $\cos^2(10)-\cos^2(20)-\cos^2(40)=(1-\sin^2(10))-(1-\sin^2(20))-(1-\sin^2(40))=-1-\left(\sin^2(10)-\sin^2(20)-\sin^2(40)\right)=-1+\frac{1}{2}$ – vnd May 19 '16 at 04:36
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    @almagest This isn't necessarily sufficient to show this, one must use the double angle theorems to reduce all of the angles to the same value. – shai horowitz May 19 '16 at 04:39

2 Answers2

5

Let me show the first identity. By the double-angle identity: \begin{align} & \sin^2(10^\circ) - \sin^2(20^\circ) - \sin^2(40^\circ) \\ = & \frac{1 - \cos(20^\circ)}{2} - \frac{1 - \cos(40^\circ)}{2} - \frac{1 - \cos(80^\circ)}{2} \\ = & -\frac{1}{2} - \frac{1}{2}(\cos(20^\circ) - \cos(40^\circ) - \cos(80^\circ)).\\ \end{align}

So to show the result, it suffices to show that $\cos(20^\circ) - \cos(40^\circ) - \cos(80^\circ) = 0$. Indeed, \begin{align} & \cos(80^\circ) = \cos(60^\circ + 20^\circ) = \frac{1}{2}\cos(20^\circ) - \frac{\sqrt{3}}{2}\sin(20^\circ), \\ & \cos(40^\circ) = \cos(60^\circ - 20^\circ) = \frac{1}{2}\cos(20^\circ) + \frac{\sqrt{3}}{2}\sin(20^\circ). \end{align} Adding these two equations gives that $\cos(20^\circ) = \cos(80^\circ) + \cos(40^\circ)$, thus the result follows.

Zhanxiong
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Let $C=\cos^2A-\cos^2B-\cos^2C$

and $S=\sin^2A-\sin^2B-\sin^2C$

$\implies C+S=-1$

and $C-S=\cos2A-\cos2B-\cos2C=\cos2A-2\cos(B+C)\cos(B-C)$ using Prosthaphaeresis Formula

If $C-S=0, C=S=-\dfrac12\ \ \ \ (0)$

Now if $B+C=60^\circ$ or more generally, $360^\circ n\pm60^\circ, \ \ \ \ (1)$

$C-S=\cos2A-\cos(B-C)$ will be $0$ if $B-C=360^\circ m\pm2A\ \ \ \ (2)$

$\implies B=180^\circ n+180^\circ m\pm30^\circ\pm A\ \ \ \ (3)$

and $C=180^\circ n-180^\circ m\pm30^\circ\mp A\ \ \ \ (4)$

Here $B=20^\circ,C=40^\circ\implies B+C=?$

and $A=10^\circ\implies B-C=360^\circ\cdot0+2A$

So, $(0)$ will hold true for $A,B,C$ satisfying $(1),(2)$

lab bhattacharjee
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