When I not seen a formula of prostapheresis in an identity We suppose that I must to proof this identity
$$\sin(220°)+\cos(10°)=\cos(70°)$$
Easily if I put $\cos(10°)$ to RHS of the identity, I can apply the formula of prostapheresis $\cos(\alpha)-\cos(\beta)$ and I solved all.

But if we not will put the $\cos(10°)$ to the RHS, the identity is it possible to solve with a calculator or is there another alternative?

 A: We can simplify as follows.
$\sin 220^{\circ} = \sin (180^{\circ}+40^{\circ}) = -\sin 40^{\circ}$
$\cos 10^{\circ}=\sin (90^{\circ}-10^{\circ})=\sin 80^{\circ}$
Then RHS becomes
$$\sin 80^{\circ}-\sin 40^{\circ}=\sin (60^{\circ}+20^{\circ})- \sin(60^{\circ}-20^{\circ}) $$
$$=2\cos 60^{\circ} \sin 20^{\circ}$$
$$=\sin 20^{\circ}=\cos 70^{\circ}$$
A: You can also have:
$\cos(10)=\sin(90-10)=\sin( 80)$
$\sin (220)+\sin(80)=2 \sin (\frac{220+80}2) \cos(\frac{220-80}2)=2 \sin(150)\cos(70)=\cos(70)$
beacause
$\sin(150)=\sin(180-30)=\sin(30)=\frac 12$
A: Firstly we need to realise that $\sin 220^\circ=\cos130^\circ$, as Blue has noted in comments. Now, using the angle addition formulae
$$\begin{align}\cos 130^\circ&=\cos(120^\circ+10^\circ)=\cos120^\circ \cos 10^\circ-\sin120^\circ\sin10^\circ\\
&=-0.5\cos10^\circ-\frac{\sqrt{3}}{2}\sin10^\circ
\end{align}$$
Hence,
$$\begin{align}
\sin220^\circ+\cos10^\circ&=\cos10^\circ-0.5\cos10^\circ-\frac{\sqrt{3}}{2}\sin10^\circ\\
&=0.5\cos10^\circ-\frac{\sqrt{3}}{2}\sin10^\circ
\end{align}$$
Now we can easily recognise this, again using the angle addition formulae, as equal to
$$\cos 60^\circ\cos10^\circ-\sin60^\circ\sin10^\circ$$
which is in fact
$$\cos(60^\circ+10^\circ)$$
and so we have:
$$\sin220^\circ+\cos10^\circ=\cos70^\circ$$
as required.
