Question

If $\cos \left(9\alpha \right)=\sin \left(\alpha \right)$ then the value of $\alpha$ is?

Answer 1

Find the required value

Given, $\cos \left(9\alpha \right)=\sin \left(\alpha \right)$ and assume that $9\alpha <90°$

That is $9\alpha$ is a acute angle

We know that,

$\sin \left(90°-\theta \right)=\cos \left(\theta \right)$

So,

$$\begin{gathered} \cos \left(9\alpha \right)=\sin \left(\alpha \right) \\ \Rightarrow \sin \left(90°-9\alpha \right)=\sin \left(\alpha \right) \\ \Rightarrow 90°-9\alpha =\alpha \\ \Rightarrow 90°=10\alpha \\ \Rightarrow \frac{90°}{10}=\alpha \\ \Rightarrow 9°=\alpha \end{gathered}$$

Hence, the value of $\alpha$ is $9°$