Question:
If $\sin 3 \theta=\cos \left(\theta-6^{\circ}\right)$, where $3 \theta$ and $\theta-6^{\circ}$ are acute angles, find the value of $\theta$.
Solution:
We have: $\sin 3 \theta=\cos \left(\theta-6^{\circ}\right)$ where $3 \theta$ and $\left(\theta-6^{\circ}\right)$ are acute angles
We have to find $\theta$
Now we proceed as to find $\theta$
$\sin 3 \theta=\cos \left(\theta-6^{\circ}\right)$
$\Rightarrow \sin 3 \theta=\sin \left[90^{\circ}-\left(\theta-6^{\circ}\right)\right]$
$\Rightarrow 3 \theta=90^{\circ}-\theta+6^{\circ}$
$\Rightarrow 4 \theta=96^{\circ}$
Therefore $\theta=24^{\circ}$