Question

If $\sin 3$ $\theta$ = $\cos$ ($\theta -6^{\circ }$), where $3$ $\theta$ and $\theta -6^{\circ }$ are acute, find the value of $\theta$.

Answer 1

Given,

$3\theta$ and $(\theta -6^{\circ })$ are acute and $\sin 3\theta =\cos (\theta -6^{\circ })$

We have, $\cos \theta =\sin ({90}^{\circ }-\theta )$

$\cos (\theta -6^{\circ })=\sin ({90}^{\circ }-(\theta -6^{\circ }\left)\right)=\sin ({96}^{\circ }-\theta )$

$\therefore \sin 3\theta =\sin ({96}^{\circ }-\theta )$

$3\theta ={96}^{\circ }-\theta$

$3\theta +\theta ={96}^{\circ }$

$4\theta ={96}^{\circ }$

$\theta =\frac{{96}^{\circ }}{4}$

$\theta ={24}^{\circ }$