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

Prove that, sinθ + 2cosθ = 1

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

Commented by mrW1 last updated on 21/Apr/17

this is not true. θ=0⇒sin θ+2cos θ=2≠1

$${this}\:{is}\:{not}\:{true}.\: \\ $$$$\theta=\mathrm{0}\Rightarrow\mathrm{sin}\:\theta+\mathrm{2cos}\:\theta=\mathrm{2}\neq\mathrm{1} \\ $$

Commented by tawa last updated on 21/Apr/17

God bless you sir.

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

Answered by mrW1 last updated on 23/Apr/17

do you mean to solve sin θ+2cos θ=1 ? (√5)((1/(√5))sinθ + (2/(√5))cosθ) = 1 (√5)(cos αsinθ + sin αcosθ) = 1 with α=sin^(−1) ((2/((√5))))) (√5)sin (θ+α)=1 sin (θ+α)=(1/(√5)) ⇒θ+α=sin^(−1) ((1/(√5))) ⇒θ=sin^(−1) ((1/(√5)))−sin^(−1) ((2/(√5))) sin θ=(1/(√5))×(1/(√5))−(2/(√5))×(2/(√5))=−(3/5) cos θ=(4/5) ⇒θ=−sin^(−1) ((3/5))+2nπ

$${do}\:{you}\:{mean}\:{to}\:{solve}\:\mathrm{sin}\:\theta+\mathrm{2cos}\:\theta=\mathrm{1}\:? \\ $$$$\sqrt{\mathrm{5}}\left(\frac{\mathrm{1}}{\sqrt{\mathrm{5}}}\mathrm{sin}\theta\:+\:\frac{\mathrm{2}}{\sqrt{\mathrm{5}}}\mathrm{cos}\theta\right)\:=\:\mathrm{1} \\ $$$$\sqrt{\mathrm{5}}\left(\mathrm{cos}\:\alpha\mathrm{sin}\theta\:+\:\mathrm{sin}\:\alpha\mathrm{cos}\theta\right)\:=\:\mathrm{1} \\ $$$${with}\:\alpha=\mathrm{sin}^{−\mathrm{1}} \:\left(\frac{\mathrm{2}}{\left.\sqrt{\mathrm{5}}\right)}\right) \\ $$$$\sqrt{\mathrm{5}}\mathrm{sin}\:\left(\theta+\alpha\right)=\mathrm{1} \\ $$$$\mathrm{sin}\:\left(\theta+\alpha\right)=\frac{\mathrm{1}}{\sqrt{\mathrm{5}}} \\ $$$$\Rightarrow\theta+\alpha=\mathrm{sin}^{−\mathrm{1}} \:\left(\frac{\mathrm{1}}{\sqrt{\mathrm{5}}}\right) \\ $$$$\Rightarrow\theta=\mathrm{sin}^{−\mathrm{1}} \:\left(\frac{\mathrm{1}}{\sqrt{\mathrm{5}}}\right)−\mathrm{sin}^{−\mathrm{1}} \:\left(\frac{\mathrm{2}}{\sqrt{\mathrm{5}}}\right) \\ $$$$\mathrm{sin}\:\theta=\frac{\mathrm{1}}{\sqrt{\mathrm{5}}}×\frac{\mathrm{1}}{\sqrt{\mathrm{5}}}−\frac{\mathrm{2}}{\sqrt{\mathrm{5}}}×\frac{\mathrm{2}}{\sqrt{\mathrm{5}}}=−\frac{\mathrm{3}}{\mathrm{5}} \\ $$$$\mathrm{cos}\:\theta=\frac{\mathrm{4}}{\mathrm{5}} \\ $$$$\Rightarrow\theta=−\mathrm{sin}^{−\mathrm{1}} \:\left(\frac{\mathrm{3}}{\mathrm{5}}\right)+\mathrm{2}{n}\pi \\ $$

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