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Question Number 126562 by mey3nipaba last updated on 21/Dec/20

$$Provethat2\sin \frac{\theta +\varphi }{2}\cos \frac{\theta -\varphi }{2}=\sin \theta +\sin \mathrm{\varnothing }$$$$Ineedhelpimmediatelyplease$$

Commented by mr W last updated on 21/Dec/20

$$\theta =\frac{\theta +\phi }{2}+\frac{\theta -\phi }{2}$$$$\phi =\frac{\theta +\phi }{2}-\frac{\theta -\phi }{2}$$$$\sin \theta =\sin (\frac{\theta +\phi }{2}+\frac{\theta -\phi }{2})=\sin \frac{\theta +\phi }{2}\cos \frac{\theta -\phi }{2}+\cos \frac{\theta +\phi }{2}\sin \frac{\theta -\phi }{2}$$$$\sin \phi =\sin (\frac{\theta +\phi }{2}-\frac{\theta -\phi }{2})=\sin \frac{\theta +\phi }{2}\cos \frac{\theta -\phi }{2}-\cos \frac{\theta +\phi }{2}\sin \frac{\theta -\phi }{2}$$$$\Rightarrow \sin \theta +\sin \phi =2\sin \frac{\theta +\phi }{2}\cos \frac{\theta -\phi }{2}$$

Answered by Olaf last updated on 21/Dec/20

$$\mathrm{Many}\mathrm{ways}\mathrm{to}\mathrm{prove}\mathrm{that}.$$$$\mathrm{Example}:$$$$\mathrm{Let}{f}_{\varphi }\left(\theta \right)=2\sin \frac{\theta +\varphi }{2}\cos \frac{\theta -\varphi }{2}$$$$\mathrm{and}{g}_{\varphi }\left(\theta \right)=\sin \theta +\sin \varphi$$$$\left(\varphi \mathrm{is}\mathrm{a}\mathrm{parameter}\right)$$$$f_{\varphi }^{\prime }\left(\theta \right)=2[\frac{1}{2}\cos \frac{\theta +\varphi }{2}\cos \frac{\theta -\varphi }{2}-\frac{1}{2}\sin \frac{\theta +\varphi }{2}\cos \frac{\theta +\varphi }{2}]$$$$f_{\varphi }^{\prime }\left(\theta \right)=\cos (\frac{\theta +\varphi }{2}+\frac{\theta -\varphi }{2})=\cos \theta =g_{\varphi }^{\prime }\left(\theta \right)$$$$\Rightarrow f_{\varphi }\left(\theta \right)=g_{\varphi }\left(\theta \right)+\mathrm{C}$$$$\mathrm{C}=f_{\varphi }\left(\varphi \right)-g_{\varphi }\left(\varphi \right)=2\sin \varphi -2\sin \varphi =0$$$$f_{\varphi }\left(\theta \right)=g_{\varphi }\left(\theta \right)$$

Answered by physicstutes last updated on 21/Dec/20

$$\mathrm{lets}\mathrm{begin}\mathrm{with}$$$$\sin (A+B)=\sin A\cos B+\sin B\cos A......\left(\mathrm{x}\right)$$$$\sin (A-B)=\sin A\cos B-\sin B\cos A.......\left(\mathrm{y}\right)$$$$\mathrm{let}A+B=\theta .....\left(i\right)$$$$\mathrm{and}A-B=\mathrm{\varnothing }.....\left(ii\right)$$$$\left(ii\right)+\left(i\right)\Rightarrow A=\frac{\theta +\mathrm{\varnothing }}{2}\mathrm{similarly}B=\frac{\theta -\mathrm{\varnothing }}{2}$$$$\left(\mathrm{x}\right)+\left(\mathrm{y}\right)\Rightarrow \sin \mathrm{\varnothing }+\sin \theta =2\sin A\cos B$$$$\Rightarrow \sin \theta +\sin \mathrm{\varnothing }=2\sin \left(\frac{\theta +\mathrm{\varnothing }}{2}\right)\cos \left(\frac{\theta -\mathrm{\varnothing }}{2}\right)$$

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