Prove this trigonometric identity...

Question

$$\sin({495}^{\circ})-\sin({795}^{\circ})+sin({1095}^{\circ})=0$$ So I have to prove that the identity is correct. How can I transform those large angles in smaller ones?

Answer 1

Using Prosthaphaeresis Formulas,

$$\sin 495^\circ+\sin1095^\circ=2\sin\frac{495^\circ+1095^\circ}2\cos\frac{1095^\circ-495^\circ}2$$

Now $\cos300^\circ=\cos(360-60)^\circ=\cos60^\circ$

Answer 2

HINT:

$$\sin1095^\circ=\sin(3\cdot360^\circ+15^\circ)=\sin15^\circ=\sin(45^\circ-30^\circ)$$

Use $\displaystyle\sin(A-B)=\sin A\cos B-\cos A\sin B$

Answer 3

$\sin(495)=\sin(135+360)=\sin(135)$; $\sin(795)=\sin(75+2(360))=\sin(75)$; $\sin(1095)=\sin(15+3(360))=\sin(15)$. With this, we have

$\sin(135)-\sin(75)+\sin(15)=0$