Question

If we have a trigonometric equation as \[\sin \left( \theta +{{23}^{0}} \right)=\cos \left( {{58}^{0}} \right)\], then what is the value of \[\cos 5\theta \] ?

Answer 1

Hint: For solving this problem first we find the value of \[\theta \] from the first equation given. We have to change all the terms in LHS and RHS into same trigonometric parameter that is we change all terms either in \[\sin \theta \] or \[\cos \theta \] by using the formulae \[\cos \theta =\sin \left( {{90}^{0}}-\theta \right)\] so that we can equate the angles directly that is if \[\sin \left( {{\theta }_{1}} \right)=\sin \left( {{\theta }_{2}} \right)\Rightarrow {{\theta }_{1}}={{\theta }_{2}}\], when \[{{\theta }_{1}} and {{\theta }_{2}}\in \left( 0,{{90}^{0}} \right)\]. Then we get \[\theta \] to find required value.

Complete step-by-step solution
Let us consider the given equation that is
\[\sin \left( \theta +{{23}^{0}} \right)=\cos \left( {{58}^{0}} \right)............equation(i)\]
We know that \[\cos \theta =\sin \left( {{90}^{0}}-\theta \right)\].
By using this formulae we can write
\[\begin{align}
  & \Rightarrow \cos \left( {{58}^{0}} \right)=\sin \left( {{90}^{0}}-{{58}^{0}} \right) \\
 & \Rightarrow \cos \left( {{58}^{0}} \right)=\sin \left( {{32}^{0}} \right) \\
\end{align}\]
By substituting the above result in the equation (i) we will get
\[\begin{align}
  & \Rightarrow \sin \left( \theta +{{23}^{0}} \right)=\cos \left( {{58}^{0}} \right) \\
 & \Rightarrow \sin \left( \theta +{{23}^{0}} \right)=\sin \left( {{32}^{0}} \right) \\
\end{align}\]
Here, we know that if \[{{\theta }_{1}} and {{\theta }_{2}}\in \left( 0,{{90}^{0}} \right)\] we can write \[\sin \left( {{\theta }_{1}} \right)=\sin \left( {{\theta }_{2}} \right)\Rightarrow {{\theta }_{1}}={{\theta }_{2}}\].
Now, by applying this theorem to above equation we will get
\[\begin{align}
  & \Rightarrow \theta +{{23}^{0}}={{32}^{0}} \\
 & \Rightarrow \theta ={{9}^{0}} \\
\end{align}\]
So, the value of \[\theta \] is \[{{9}^{0}}\].
Now, let us find the value of \[\cos 5\theta \], by substituting the value of \[\theta \] we get
\[\begin{align}
  & \Rightarrow \cos 5\theta =\cos \left( 5\times {{9}^{0}} \right) \\
 & \Rightarrow \cos 5\theta =\cos \left( {{45}^{0}} \right) \\
\end{align}\]
We know that the value of \[\cos {{45}^{0}}\] is \[\dfrac{1}{\sqrt{2}}\], this is the standard value.
So by substituting this value in above equation we will get
\[\Rightarrow \cos 5\theta =\dfrac{1}{\sqrt{2}}\]
Therefore, the value of \[\cos 5\theta \] is \[\dfrac{1}{\sqrt{2}}\].

Note: We can solve this problem in different methods also.
Above we converted all terms into \[\sin \theta \], now let us convert them to \[\cos \theta \].
We know that \[\cos \theta =\sin \left( {{90}^{0}}-\theta \right)\], similarly we can write \[\sin \theta =\cos \left( {{90}^{0}}-\theta \right)\]
By taking the equation we will get
\[\begin{align}
  & \Rightarrow \sin \left( \theta +{{23}^{0}} \right)=\cos \left( {{58}^{0}} \right) \\
 & \Rightarrow \cos \left( {{90}^{0}}-\left( \theta +{{23}^{0}} \right) \right)=\cos \left( {{58}^{0}} \right) \\
 & \Rightarrow \cos \left( {{67}^{0}}-\theta \right)=\cos \left( {{58}^{0}} \right) \\
\end{align}\]
We know that \[\cos \left( {{\theta }_{1}} \right)=\cos \left( {{\theta }_{2}} \right)\Rightarrow {{\theta }_{1}}={{\theta }_{2}}\], when\[{{\theta }_{1}}and{{\theta }_{2}}\in \left( 0,{{90}^{0}} \right)\]
By applying this theorem we get
\[\begin{align}
  & \Rightarrow \left( {{67}^{0}}-\theta \right)={{58}^{0}} \\
 & \Rightarrow \theta ={{9}^{0}} \\
\end{align}\]
Now, let us find the value of \[\cos 5\theta \], by substituting the value of \[\theta \] we get
\[\begin{align}
  & \Rightarrow \cos 5\theta =\cos \left( 5\times {{9}^{0}} \right) \\
 & \Rightarrow \cos 5\theta =\cos \left( {{45}^{0}} \right) \\
\end{align}\]
We know that the value of \[\cos {{45}^{0}}\] is \[\dfrac{1}{\sqrt{2}}\], this is the standard value.
So by substituting this value in the above equation we will get
\[\Rightarrow \cos 5\theta =\dfrac{1}{\sqrt{2}}\]
Therefore, the value of \[\cos 5\theta \] is \[\dfrac{1}{\sqrt{2}}\].