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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
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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.
Last updated date: 27th Sep 2023
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