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Express-sin-3-cos-in-the-form-Rsin-where-R-gt-0-and-0-lt-lt-90-Hence-solve-the-equation-sin-3-cos-2-for-0-lt-lt-270-




Question Number 8273 by lepan last updated on 05/Oct/16
Express sinα+(√3)cosα in the form   Rsin(α+β) where R>0 and 0°<β<90°.  Hence solve the equation sinα+(√3)cosα=2  for 0°<α<270°.
$${Express}\:{sin}\alpha+\sqrt{\mathrm{3}}{cos}\alpha\:{in}\:{the}\:{form}\: \\ $$$${Rsin}\left(\alpha+\beta\right)\:{where}\:{R}>\mathrm{0}\:{and}\:\mathrm{0}°<\beta<\mathrm{90}°. \\ $$$${Hence}\:{solve}\:{the}\:{equation}\:{sin}\alpha+\sqrt{\mathrm{3}}{cos}\alpha=\mathrm{2} \\ $$$${for}\:\mathrm{0}°<\alpha<\mathrm{270}°. \\ $$
Answered by prakash jain last updated on 05/Oct/16
sin α+(√3)cos α  =2((1/2)sin α+((√3)/2)cos α)  =2(sin (π/6)sin α+cos (π/6)cos α)  =2cos ((π/6)−α)  sin α+(√3)cos α=1  ⇒2cos ((π/6)−α)=1  ⇒cos ((π/6)−α)=(1/2)  ⇒cos ((π/6)−α)=cos (π/3)  (π/6)−α=2nπ±(π/3)  α=−2nπ+((π/6)±(π/3))= { ((2kπ+(π/2))),((2kπ−(π/6))) :}  between 0 to 270° (or 0 to ((3π)/2))  α=(π/2)or 90°
$$\mathrm{sin}\:\alpha+\sqrt{\mathrm{3}}\mathrm{cos}\:\alpha \\ $$$$=\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:\alpha+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{cos}\:\alpha\right) \\ $$$$=\mathrm{2}\left(\mathrm{sin}\:\frac{\pi}{\mathrm{6}}\mathrm{sin}\:\alpha+\mathrm{cos}\:\frac{\pi}{\mathrm{6}}\mathrm{cos}\:\alpha\right) \\ $$$$=\mathrm{2cos}\:\left(\frac{\pi}{\mathrm{6}}−\alpha\right) \\ $$$$\mathrm{sin}\:\alpha+\sqrt{\mathrm{3}}\mathrm{cos}\:\alpha=\mathrm{1} \\ $$$$\Rightarrow\mathrm{2cos}\:\left(\frac{\pi}{\mathrm{6}}−\alpha\right)=\mathrm{1} \\ $$$$\Rightarrow\mathrm{cos}\:\left(\frac{\pi}{\mathrm{6}}−\alpha\right)=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{cos}\:\left(\frac{\pi}{\mathrm{6}}−\alpha\right)=\mathrm{cos}\:\frac{\pi}{\mathrm{3}} \\ $$$$\frac{\pi}{\mathrm{6}}−\alpha=\mathrm{2}{n}\pi\pm\frac{\pi}{\mathrm{3}} \\ $$$$\alpha=−\mathrm{2}{n}\pi+\left(\frac{\pi}{\mathrm{6}}\pm\frac{\pi}{\mathrm{3}}\right)=\begin{cases}{\mathrm{2}{k}\pi+\frac{\pi}{\mathrm{2}}}\\{\mathrm{2}{k}\pi−\frac{\pi}{\mathrm{6}}}\end{cases} \\ $$$$\mathrm{between}\:\mathrm{0}\:\mathrm{to}\:\mathrm{270}°\:\left({or}\:\mathrm{0}\:\mathrm{to}\:\frac{\mathrm{3}\pi}{\mathrm{2}}\right) \\ $$$$\alpha=\frac{\pi}{\mathrm{2}}\mathrm{or}\:\mathrm{90}° \\ $$

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