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

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

Explanation

Solution

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

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

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

∴\therefore The answer is sin⁡α\sin \alpha .

Note: We have other trigonometric identities.
For some angle AA,
sin⁡2A=2tan⁡A1+tan⁡2A\sin 2A = \dfrac{{2\tan A}}{{1 + {{\tan }^2}A}}
cos⁡2A=2cos⁡2A−1\cos 2A = 2{\cos ^2}A - 1
cos⁡2A=cos⁡2A−sin⁡2A\cos 2A = {\cos ^2}A - {\sin ^2}A
Like these, we have other identities for sin⁡,cos⁡,tan⁡,cosec,sec\sin ,\cos ,\tan ,{\text{cosec,sec}} and cot⁡\cot .