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

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