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.