Question #02a6f

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Answer 1 · 2017-02-06T14:02:36

To be contd.............

We use, $a^3+b^3=(a+b)(a^2-ab+b^2)$ to get,

$(costheta)^6+(sintheta)^6=cos^6theta+sin^6theta$

$=(cos^2theta+sin^2theta)(cos^4theta-cos^2thetasin^2theta+sin^4theta)$

$=(1){(cos^2theta+sin^2theta)^2-2sin^2thetacos^2theta-cos^2thetasin^2theta}$

$=1-3cos^2thetasin^2theta$

$=1-3/4(2sinthetacostheta)^2$

$=1-3/4(sin2theta)^2$

$=1-3/4(sin^2(2theta))$

Recall that, $1-cos2A=2sin^2A$ .

Hence, the Exp. $=1-3/8{2sin^2(2theta)}$

$=1-3/8(1-cos4theta)$

$=1-3/8+3/8cos4theta$

$=5/8+3/8cos4theta$