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


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Question Number 43266 by gunawan last updated on 09/Sep/18

$\cos^{3} x + \cos^{- 3} x = 0$ $\sin 2 x + \cos 2 x = . . .$
	Answered by tanmay.chaudhury50@gmail.com last updated on 09/Sep/18
	$t=cosx+isinx=e^(ix) (1/t)=cosx−isinx=e^(−ix) t+(1/t)=2cosx (t+(1/t))^3 =2^3 cos^3 x (1/2^3 )(t+(1/t))^3 =cos^3 x now (1/2^3 )(t+(1/t))^3 +(1/((1/2^3 )×(t+(1/t))^3 ))=0 {(1/2^3 )(t+(1/t))^3 }^2 +1=0 (1/2^6 )(t+(1/t))^6 +1=0 now (t+(1/t))^6 +2^6 =0 (t+(1/t))^6 =2^6 i^6 t+(1/t)=2i 2cosx=2i cosx=i sin2x+cos2x =2sinx.cosx+2cos^2 x−1 =2(√(1−cos^2 x)) cosx+2cos^2 x−1 =2(√(1−i^2 )) ×i+2i^2 −1 =2(√2) ×i+2(−1)−1 =−3+2(√2) i t$
	$t = c o s x + i s i n x = e^{i x}$ $\frac{1}{t} = c o s x - i s i n x = e^{- i x}$ $t + \frac{1}{t} = 2 c o s x$ ${(t + \frac{1}{t})}^{3} = 2^{3} {c o s}^{3} x$ $\frac{1}{2^{3}} {(t + \frac{1}{t})}^{3} = {c o s}^{3} x$ $n o w$ $\frac{1}{2^{3}} {(t + \frac{1}{t})}^{3} + \frac{1}{\frac{1}{2^{3}} \times {(t + \frac{1}{t})}^{3}} = 0$ ${\frac{1}{2^{3}} {(t + \frac{1}{t})}^{3}}^{2} + 1 = 0$ $\frac{1}{2^{6}} {(t + \frac{1}{t})}^{6} + 1 = 0$ $n o w$ ${(t + \frac{1}{t})}^{6} + 2^{6} = 0$ ${(t + \frac{1}{t})}^{6} = 2^{6} i^{6}$ $t + \frac{1}{t} = 2 i$ $2 c o s x = 2 i$ $c o s x = i$ $s i n 2 x + c o s 2 x$ $= 2 s i n x . c o s x + 2 {c o s}^{2} x - 1$ $= 2 \sqrt{1 - {c o s}^{2} x} c o s x + 2 {c o s}^{2} x - 1$ $= 2 \sqrt{1 - i^{2}} \times i + 2 i^{2} - 1$ $= 2 \sqrt{2} \times i + 2 (- 1) - 1$ $= - 3 + 2 \sqrt{2} i$ $t$
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