Question

If $\sin 3\theta = \cos \left( {\theta - {6^0}} \right)$, where $3\theta$ and $\left( {\theta - {6^0}} \right)$ are acute, find the value of $\theta$.

Answer 1

Hint- Here, we will be using the trigonometric function $\sin \phi = \cos \left( {{{90}^0} - \phi } \right)$.

Given, $\sin 3\theta = \cos \left( {\theta - {6^0}} \right){\text{ }} \to {\text{(1)}}$
We know that $\sin \phi = \cos \left( {{{90}^0} - \phi } \right)$ where $\phi$ is an acute angle.
As, $3\theta$ is also acute angle so we can write $\sin 3\theta = \cos \left( {{{90}^0} - 3\theta } \right)$
Therefore, equation (1) becomes

$$\Rightarrow \cos \left( {{{90}^0} - 3\theta } \right) = \cos \left( {\theta - {6^0}} \right) \Rightarrow {90^0} - 3\theta = \theta - {6^0} \Rightarrow 4\theta = {96^0} \\\ \Rightarrow \theta = {24^0} \\\$$

Further also we have to check whether the angles $3\theta$ and $\left( {\theta - {6^0}} \right)$ are
coming acute angles or not.
For $\theta = {24^0}$, $3\theta = {72^0}$ and $\left( {\theta - {6^0}} \right) = {18^0}$
That means both the angles are coming acute so $\theta = {24^0}$ which is the required acute angle.

Note- In these types of problems, we convert both the LHS and the RHS of the given equation into one
trigonometric function and then compare the angles to solve for the unknown.