<s>[INST] 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.
 [/INST] 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.
</s>