# $\sin (y-30^{\circ})=\cos y, 0^{\circ}\leq y \leq360^{\circ}$

Given that $$\sin (y-30^{\circ})=\cos y, 0^{\circ}\leq y \leq360^{\circ}$$

My attempt,

$\sin y \cos(-30^{\circ})-\cos y \sin (-30^{\circ})=\cos y$

$\frac{\sqrt3}{2}\sin y+\frac{1}{2}\cos y=\cos y$

$\frac{\sqrt3}{2}\sin y=\frac{1}{2}\cos y$

$\frac{\sqrt3}{2}\tan y=\frac{1}{2}$

$\tan y=\frac{\sqrt 3}{3}$

$y=30^{\circ},210^{\circ}$

But the answers are incorrect. Anything wrong with my solution?

• There is a typo : $\frac{\color{red}{\sqrt3}}{2}\tan y=\frac{1}{2}$, corrected – Jaideep Khare May 5 '17 at 22:51
• What answer is given there in your book? – Jaideep Khare May 5 '17 at 22:53
• 60 degree and 240 degree – Mathxx May 5 '17 at 22:56

The problem is , you expanded :

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

Which is definitely wrong

You either expand it as : $$\sin(A-B)=\sin(A)\cos(B)-\cos(A)\sin(B)$$ or $$\sin(A-B)=\sin(A)\cos(-B)+\cos(A)\sin(-B)$$

But not both simultaneously.

Hint:

It's much simpler to turn the sine into a cosine using the formulae for complementary angles:

$\sin(y-30)=\cos(90-(y-30))=\cos(120-y)$, so you have to solve the simpler equation: $$\cos(120-y)=\cos y.$$

• OP is asking Anything wrong with my solution ? not any alternate way. – Jaideep Khare May 5 '17 at 23:00
• The answer could have started with "avoid expansion & use $$\sin(90^\circ-y)=\cos y$$ or $$\cos(90^\circ-y)=\sin y$$" – lab bhattacharjee May 6 '17 at 1:49
• @Iab bhattacharjee: You're right. I'll fix that. – Bernard May 6 '17 at 9:03