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Question Number 115765 by aye48 last updated on 28/Sep/20

$$$$$$\mathrm{Show}\mathrm{that}\sin (\alpha +\beta )=\sin \alpha \cos \beta +\cos \alpha \sin \beta .$$

Commented by bemath last updated on 28/Sep/20

Commented by aye48 last updated on 28/Sep/20

$$\mathrm{Tk}\mathrm{yu}\mathrm{sir},\mathrm{explain}\mathrm{step}\mathrm{by}\mathrm{step}\mathrm{pl}.$$

Commented by Ar Brandon last updated on 28/Sep/20

$$\sin (\alpha +\beta )=\frac{\mathrm{PT}}{\mathrm{OP}}=\frac{\mathrm{RT}+\mathrm{PR}}{\mathrm{OP}}=\frac{\mathrm{SU}+\mathrm{PR}}{\mathrm{OP}}$$$$=\frac{\mathrm{SU}}{\mathrm{OP}}+\frac{\mathrm{PR}}{\mathrm{OP}}=\frac{\mathrm{SU}}{\mathrm{OS}}\times \frac{\mathrm{OS}}{\mathrm{OP}}+\frac{\mathrm{PR}}{\mathrm{PS}}\times \frac{\mathrm{PS}}{\mathrm{OP}}$$$$=\mathbf{sin}\mathit{\alpha }\mathbf{cos}\mathit{\beta }+\mathbf{cos}\mathit{\alpha }\mathbf{sin}\mathit{\beta }$$

Answered by MJS_new last updated on 28/Sep/20

$$\mathrm{remember}$$$$\sin x=\frac{{\mathrm{e}}^{\mathrm{i}x}-{\mathrm{e}}^{-\mathrm{i}x}}{2\mathrm{i}}=-\frac{\mathrm{i}}{2}({\mathrm{e}}^{\mathrm{i}x}-\frac{1}{{\mathrm{e}}^{\mathrm{i}x}})$$$$\cos x=\frac{{\mathrm{e}}^{\mathrm{u}x}+{\mathrm{e}}^{-\mathrm{i}x}}{2}$$$$\Rightarrow$$$$\sin (\alpha +\beta )=-\frac{\mathrm{i}}{2}({\mathrm{e}}^{\mathrm{i}\alpha }{\mathrm{e}}^{\mathrm{i}\beta }-\frac{1}{{\mathrm{e}}^{\mathrm{i}\alpha }{\mathrm{e}}^{\mathrm{i}\beta }})=$$$$[\mathrm{let}{\mathrm{e}}^{\mathrm{i}\alpha }=p\wedge {\mathrm{e}}^{\mathrm{i}\beta }=q]$$$$=-\frac{\mathrm{i}}{2}(pq-\frac{1}{pq})=$$$$[\mathrm{now}\mathrm{the}\mathrm{trick}:]$$$$=-\frac{\mathrm{i}}{4}(2pq-\frac{2}{pq})=$$$$=-\frac{\mathrm{i}}{4}(2pq-\frac{2}{pq}+(\frac{p}{q}+\frac{q}{p})-(\frac{p}{q}+\frac{q}{p}))=$$$$=-\frac{\mathrm{i}}{4}((pq+\frac{p}{q}-\frac{q}{p}-\frac{1}{pq})+(pq-\frac{p}{q}+\frac{q}{p}-\frac{1}{pq}))=$$$$=-\frac{\mathrm{i}}{4}((p-\frac{1}{p})(q+\frac{1}{q})+(p+\frac{1}{p})(q-\frac{1}{q}))=$$$$=-\frac{\mathrm{i}(p-\frac{1}{p})}{2}\times \frac{q+\frac{1}{q}}{2}-\frac{p+\frac{1}{p}}{2}\times \frac{\mathrm{i}(q-\frac{1}{q})}{2}=$$$$=\frac{{\mathrm{e}}^{\mathrm{i}\alpha }-{\mathrm{e}}^{-\mathrm{i}\alpha }}{2\mathrm{i}}\times \frac{{\mathrm{e}}^{\mathrm{i}\beta }+{\mathrm{e}}^{-\mathrm{i}\beta }}{2}+\frac{{\mathrm{e}}^{\mathrm{i}\alpha }+{\mathrm{e}}^{-\mathrm{i}\alpha }}{2}\times \frac{{\mathrm{e}}^{\mathrm{i}\beta }-{\mathrm{e}}^{-\mathrm{i}\beta }}{2\mathrm{i}}=$$$$=\sin \alpha \cos \beta +\cos \alpha \sin \beta$$

Commented by Dwaipayan Shikari last updated on 28/Sep/20

$$\mathrm{Great}\mathrm{thinking}\mathrm{sir}!$$

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