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Question Number 12378 by tawa last updated on 20/Apr/17

$$\mathrm{Prove}\mathrm{that},$$$$\sin \theta +2\cos \theta =1$$

Commented by mrW1 last updated on 21/Apr/17

$$thisisnottrue.$$$$\theta =0\Rightarrow \sin \theta +2\cos \theta =2\ne 1$$

Commented by tawa last updated on 21/Apr/17

$$\mathrm{God}\mathrm{bless}\mathrm{you}\mathrm{sir}.$$

Answered by mrW1 last updated on 23/Apr/17

$$doyoumeantosolve\sin \theta +2\cos \theta =1?$$$$\sqrt{5}(\frac{1}{\sqrt{5}}\sin \theta +\frac{2}{\sqrt{5}}\cos \theta )=1$$$$\sqrt{5}(\cos \alpha \sin \theta +\sin \alpha \cos \theta )=1$$$$with\alpha ={\sin }^{-1}\left(\frac{2}{\sqrt{5})}\right)$$$$\sqrt{5}\sin (\theta +\alpha )=1$$$$\sin (\theta +\alpha )=\frac{1}{\sqrt{5}}$$$$\Rightarrow \theta +\alpha ={\sin }^{-1}\left(\frac{1}{\sqrt{5}}\right)$$$$\Rightarrow \theta ={\sin }^{-1}\left(\frac{1}{\sqrt{5}}\right)-{\sin }^{-1}\left(\frac{2}{\sqrt{5}}\right)$$$$\sin \theta =\frac{1}{\sqrt{5}}\times \frac{1}{\sqrt{5}}-\frac{2}{\sqrt{5}}\times \frac{2}{\sqrt{5}}=-\frac{3}{5}$$$$\cos \theta =\frac{4}{5}$$$$\Rightarrow \theta =-{\sin }^{-1}\left(\frac{3}{5}\right)+2n\pi$$

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