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

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

& \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{aligned}$$ By substituting the above result in the equation (i) we will get $$\begin{aligned} & \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{aligned}$$ Here, we know that if $${{\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{aligned} & \Rightarrow \theta +{{23}^{0}}={{32}^{0}} \\\ & \Rightarrow \theta ={{9}^{0}} \\\ \end{aligned}$$ So, the value of $$\theta $$ is $${{9}^{0}}$$. Now, let us find the value of $$\cos 5\theta $$, by substituting the value of $$\theta $$ we get $$\begin{aligned} & \Rightarrow \cos 5\theta =\cos \left( 5\times {{9}^{0}} \right) \\\ & \Rightarrow \cos 5\theta =\cos \left( {{45}^{0}} \right) \\\ \end{aligned}$$ We know that the value of $$\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{aligned} & \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{aligned}$$ We know that $$\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{aligned} & \Rightarrow \left( {{67}^{0}}-\theta \right)={{58}^{0}} \\\ & \Rightarrow \theta ={{9}^{0}} \\\ \end{aligned}$$ Now, let us find the value of $$\cos 5\theta $$, by substituting the value of $$\theta $$ we get $$\begin{aligned} & \Rightarrow \cos 5\theta =\cos \left( 5\times {{9}^{0}} \right) \\\ & \Rightarrow \cos 5\theta =\cos \left( {{45}^{0}} \right) \\\ \end{aligned}$$ We know that the value of $$\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}}$$.