Question and Answers Forum

All Questions      Topic List

Trigonometry Questions

Previous in All Question      Next in All Question      

Previous in Trigonometry      Next in Trigonometry      

Question Number 8306 by lepan last updated on 07/Oct/16

If 270°<x<360°, simplify (√(2+(√(2+2cosx)))).

$${If}\:\mathrm{270}°<{x}<\mathrm{360}°,\:{simplify} \\ $$ $$\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\mathrm{2}{cosx}}}. \\ $$

Commented bysou1618 last updated on 07/Oct/16

(√(2+(√(2+2cosx)))) (270°<x<360°) 2cos^2 α=1+cos2α =(√(2+(√(4cos^2 ((1/2)x))))) =(√(2+∣2cos((1/2)x)∣)) 135°<(1/2)x<180° ⇒ cos((1/2)x)<0 =(√(2−2cos((1/2)x))) 2sin^2 α=1−cos2α =(√(4sin^2 ((1/4)x))) =∣2sin((1/4)x)∣ 67.5°<(1/4)x<90° ⇒ sin((1/4)x)>0 =2sin(1/4)x

$$\sqrt{\mathrm{2}+\sqrt{\mathrm{2}+\mathrm{2}{cosx}}}\:\left(\mathrm{270}°<{x}<\mathrm{360}°\right) \\ $$ $$\mathrm{2}{cos}^{\mathrm{2}} \alpha=\mathrm{1}+{cos}\mathrm{2}\alpha \\ $$ $$=\sqrt{\mathrm{2}+\sqrt{\mathrm{4}{cos}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{2}}{x}\right)}} \\ $$ $$=\sqrt{\mathrm{2}+\mid\mathrm{2}{cos}\left(\frac{\mathrm{1}}{\mathrm{2}}{x}\right)\mid} \\ $$ $$\mathrm{135}°<\frac{\mathrm{1}}{\mathrm{2}}{x}<\mathrm{180}°\:\Rightarrow\:{cos}\left(\frac{\mathrm{1}}{\mathrm{2}}{x}\right)<\mathrm{0} \\ $$ $$=\sqrt{\mathrm{2}−\mathrm{2}{cos}\left(\frac{\mathrm{1}}{\mathrm{2}}{x}\right)} \\ $$ $$\mathrm{2}{sin}^{\mathrm{2}} \alpha=\mathrm{1}−{cos}\mathrm{2}\alpha \\ $$ $$=\sqrt{\mathrm{4}{sin}^{\mathrm{2}} \left(\frac{\mathrm{1}}{\mathrm{4}}{x}\right)} \\ $$ $$=\mid\mathrm{2}{sin}\left(\frac{\mathrm{1}}{\mathrm{4}}{x}\right)\mid \\ $$ $$\mathrm{67}.\mathrm{5}°<\frac{\mathrm{1}}{\mathrm{4}}{x}<\mathrm{90}°\:\Rightarrow\:{sin}\left(\frac{\mathrm{1}}{\mathrm{4}}{x}\right)>\mathrm{0} \\ $$ $$=\mathrm{2}{sin}\frac{\mathrm{1}}{\mathrm{4}}{x} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com