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 43266 by gunawan last updated on 09/Sep/18

cos^3 x+cos^(−3) x=0 sin 2x+cos 2x=...

$$\mathrm{cos}^{\mathrm{3}} {x}+\mathrm{cos}^{−\mathrm{3}} {x}=\mathrm{0} \\ $$$$\mathrm{sin}\:\mathrm{2}{x}+\mathrm{cos}\:\mathrm{2}{x}=... \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 09/Sep/18

t=cosx+isinx=e^(ix) (1/t)=cosx−isinx=e^(−ix) t+(1/t)=2cosx (t+(1/t))^3 =2^3 cos^3 x (1/2^3 )(t+(1/t))^3 =cos^3 x now (1/2^3 )(t+(1/t))^3 +(1/((1/2^3 )×(t+(1/t))^3 ))=0 {(1/2^3 )(t+(1/t))^3 }^2 +1=0 (1/2^6 )(t+(1/t))^6 +1=0 now (t+(1/t))^6 +2^6 =0 (t+(1/t))^6 =2^6 i^6 t+(1/t)=2i 2cosx=2i cosx=i sin2x+cos2x =2sinx.cosx+2cos^2 x−1 =2(√(1−cos^2 x)) cosx+2cos^2 x−1 =2(√(1−i^2 )) ×i+2i^2 −1 =2(√2) ×i+2(−1)−1 =−3+2(√2) i t

$${t}={cosx}+{isinx}={e}^{{ix}} \\ $$$$\frac{\mathrm{1}}{{t}}={cosx}−{isinx}={e}^{−{ix}} \\ $$$${t}+\frac{\mathrm{1}}{{t}}=\mathrm{2}{cosx} \\ $$$$\left({t}+\frac{\mathrm{1}}{{t}}\right)^{\mathrm{3}} =\mathrm{2}^{\mathrm{3}} {cos}^{\mathrm{3}} {x} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }\left({t}+\frac{\mathrm{1}}{{t}}\right)^{\mathrm{3}} ={cos}^{\mathrm{3}} {x} \\ $$$${now} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }\left({t}+\frac{\mathrm{1}}{{t}}\right)^{\mathrm{3}} +\frac{\mathrm{1}}{\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }×\left({t}+\frac{\mathrm{1}}{{t}}\right)^{\mathrm{3}} }=\mathrm{0} \\ $$$$\left\{\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{3}} }\left({t}+\frac{\mathrm{1}}{{t}}\right)^{\mathrm{3}} \right\}^{\mathrm{2}} +\mathrm{1}=\mathrm{0} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{6}} }\left({t}+\frac{\mathrm{1}}{{t}}\right)^{\mathrm{6}} +\mathrm{1}=\mathrm{0} \\ $$$${now} \\ $$$$\left({t}+\frac{\mathrm{1}}{{t}}\right)^{\mathrm{6}} +\mathrm{2}^{\mathrm{6}} =\mathrm{0} \\ $$$$\left({t}+\frac{\mathrm{1}}{{t}}\right)^{\mathrm{6}} =\mathrm{2}^{\mathrm{6}} {i}^{\mathrm{6}} \\ $$$${t}+\frac{\mathrm{1}}{{t}}=\mathrm{2}{i} \\ $$$$\mathrm{2}{cosx}=\mathrm{2}{i} \\ $$$${cosx}={i} \\ $$$${sin}\mathrm{2}{x}+{cos}\mathrm{2}{x} \\ $$$$=\mathrm{2}{sinx}.{cosx}+\mathrm{2}{cos}^{\mathrm{2}} {x}−\mathrm{1} \\ $$$$=\mathrm{2}\sqrt{\mathrm{1}−{cos}^{\mathrm{2}} {x}}\:\:{cosx}+\mathrm{2}{cos}^{\mathrm{2}} {x}−\mathrm{1} \\ $$$$=\mathrm{2}\sqrt{\mathrm{1}−{i}^{\mathrm{2}} }\:\:×{i}+\mathrm{2}{i}^{\mathrm{2}} −\mathrm{1} \\ $$$$=\mathrm{2}\sqrt{\mathrm{2}}\:×{i}+\mathrm{2}\left(−\mathrm{1}\right)−\mathrm{1} \\ $$$$=−\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\:{i} \\ $$$$ \\ $$$${t} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com