Question

If $\sin \theta =\cos (\theta -6^0)$, find the value of $\theta$

Answer 1

$\sin \theta =\cos (\theta -6^{\circ })$

$\Rightarrow \cos (90-\theta )=\cos (\theta -6^{\circ })$ since $\cos (90-\theta )=\sin \theta$

$\Rightarrow 90-\theta =\theta -6^{\circ }$

$\Rightarrow -2\theta =-{96}^{\circ }$

$\therefore \theta =\frac{{96}^{\circ }}{2}={48}^{\circ }$