When I not seen a formula of prostapheresis in an identity

Question

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?

Answer 1

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 $$\sin 80^{\circ}-\sin 40^{\circ}=\sin (60^{\circ}+20^{\circ})- \sin(60^{\circ}-20^{\circ}) $$ $$=2\cos 60^{\circ} \sin 20^{\circ}$$ $$=\sin 20^{\circ}=\cos 70^{\circ}$$

Answer 2

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

$\sin(150)=\sin(180-30)=\sin(30)=\frac 12$

Answer 3

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.