Question

If $\cos (\alpha + \beta ) = 0$, then find $\sin (\alpha + 2\beta )$.

Answer 1

Hint: There are trigonometric identities for $\sin (A + B)$ and $\cos (A + B)$. Using the formula of $\sin (A + B)$ we can expand the question. Then substitute using formulas of $\sin 2A$ and $\cos 2A$. Then we can simplify and substitute for $\cos (\alpha + \beta ) = 0$, we get the answer.

Formula used: For any angles $A,B$ we have these trigonometric relations.
$\cos (A + B) = \cos A\cos B - \sin A\sin B$
$\sin (A + B) = \sin A\cos B + \cos A\sin B$
$\sin 2A = 2\sin A\cos A$
$\cos 2A = 1 - 2{\sin ^2}A$

Complete step-by-step answer:
Given $\cos (\alpha + \beta ) = 0$.
We are asked to find $\sin (\alpha + 2\beta )$.
We know that,
$\sin (A + B) = \sin A\cos B + \cos A\sin B$
Let, $A = \alpha ,B = 2\beta $
So we have,
$\sin (\alpha + 2\beta ) = \sin \alpha \cos 2\beta + \cos \alpha \sin 2\beta ----- (i)$
Also we have the relations,
$\cos 2A = 1 - 2{\sin ^2}A$
$\sin 2A = 2\sin A\cos A$
This gives, $\cos 2\beta = 1 - 2{\sin ^2}\beta $ and $\sin 2\beta = 2\sin \beta \cos \beta $
Substituting these, we have from equation $(i)$,
$\Rightarrow$$\sin (\alpha + 2\beta ) = \sin \alpha (1 - 2{\sin ^2}\beta ) + \cos \alpha (2\sin \beta \cos \beta )$
Expanding the above equation we get,
$\Rightarrow$$\sin (\alpha + 2\beta ) = \sin \alpha - 2\sin \alpha {\sin ^2}\beta + 2\cos \alpha \sin \beta \cos \beta $
Taking $\sin \beta $ common from right hand side we get,
$\Rightarrow$$\sin (\alpha + 2\beta ) = \sin \alpha + 2\sin \beta (\cos \alpha \cos \beta - \sin \alpha \sin \beta ) ------ (ii)$
Now, $\cos (A + B) = \cos A\cos B - \sin A\sin B$
Using this we can write equation $(ii)$ as,
$\Rightarrow$$\sin (\alpha + 2\beta ) = \sin \alpha + 2\sin \beta \cos (\alpha + \beta )$
But in the question it is given that,
$\cos (\alpha + \beta ) = 0$
Substituting this we have,
$\Rightarrow$$\sin (\alpha + 2\beta ) = \sin \alpha + 2\sin \beta \times 0$
$ \Rightarrow \sin (\alpha + 2\beta ) = \sin \alpha + 0$
So we have,
$\sin (\alpha + 2\beta ) = \sin \alpha $

$\therefore $ The answer is $\sin \alpha $.

Note: We have other trigonometric identities.
For some angle $A$,
$\sin 2A = \dfrac{{2\tan A}}{{1 + {{\tan }^2}A}}$
$\cos 2A = 2{\cos ^2}A - 1$
$\cos 2A = {\cos ^2}A - {\sin ^2}A$
Like these, we have other identities for $\sin ,\cos ,\tan ,{\text{cosec,sec}}$ and $\cot $.