Question

# If $\sin \theta =\cos \left( \theta -{{6}^{\circ }} \right)$, find the value of $\theta$.

Hint: Use the trigonometric formula $\sin \theta =\cos \left( {{90}^{\circ }}-\theta \right)$ to simplify the given trigonometric expression. Rearrange the terms of the equation and then simplify it to find the value of angle which satisfies the given trigonometric equation.

We have the equation $\sin \theta =\cos \left( \theta -{{6}^{\circ }} \right)$. We have to find the value of $\theta$ which satisfies the given equation.
We know that $\sin \theta =\cos \left( {{90}^{\circ }}-\theta \right)$.
Substituting the above equation in equation $\sin \theta =\cos \left( \theta -{{6}^{\circ }} \right)$, we have $\cos \left( {{90}^{\circ }}-\theta \right)=\cos \left( \theta -{{6}^{\circ }} \right)$.
Thus, we have ${{90}^{\circ }}-\theta =\theta -{{6}^{\circ }}$.
Rearranging the terms, we have $2\theta ={{90}^{\circ }}+{{6}^{\circ }}={{96}^{\circ }}$.
Thus, we have $\theta =\dfrac{{{96}^{\circ }}}{2}={{48}^{\circ }}$.
Hence, the value of $\theta$ which satisfies the given equation is $\theta ={{48}^{\circ }}$.
Note: We can also solve this question by using the identity $\cos \left( x-y \right)=\cos x\cos y+\sin x\sin y$. Substitute $x=\theta ,y={{6}^{\circ }}$ in the identity and simplify it to get $\tan \theta =\dfrac{\cos {{6}^{\circ }}}{1-\sin {{6}^{\circ }}}$. Substitute the values $\cos {{6}^{\circ }}=\dfrac{1}{4}\sqrt{7+\sqrt{5}+\sqrt{30+6\sqrt{5}}}$ and $\sin {{6}^{\circ }}=\dfrac{\sqrt{9-\sqrt{5}-\sqrt{30+6\sqrt{5}}}}{4}$ in the previous equation. Also, use the fact that ${{\tan }^{-1}}\left( 1.11 \right)={{48}^{\circ }}$ to get the value of $\theta$ which satisfies the given equation.