# 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 $.

Last updated date: 27th Mar 2023

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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.

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.

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