# If sin 3 θ = cos (θ − 6°),

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 $\theta=24^{\circ}$