Prove $\sin^2(10 ^\circ)-\sin^2(20^\circ)-\sin^2(40^\circ)=-\frac{1}{2}$ identity 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.
 A: 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.
A: 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 $(1),(2)$
