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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?

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

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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: $$\cos(120-y)=\cos y.$$

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

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