Question #02a6f

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Question #02a6f

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Answer 1 · 2017-02-06T14:02:36

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

Explanation:

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

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

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

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

$= 1 - 3 {cos}^{2} θ {sin}^{2} θ$ $=1-3cos^2thetasin^2theta$

$= 1 - \frac{3}{4} {(2 sin θ cos θ)}^{2}$ $=1-3/4(2sinthetacostheta)^2$

$= 1 - \frac{3}{4} {(sin 2 θ)}^{2}$ $=1-3/4(sin2theta)^2$

$= 1 - \frac{3}{4} ({sin}^{2} (2 θ))$ $=1-3/4(sin^2(2theta))$

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

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

$= 1 - \frac{3}{8} (1 - cos 4 θ)$ $=1-3/8(1-cos4theta)$

$= 1 - \frac{3}{8} + \frac{3}{8} cos 4 θ$ $=1-3/8+3/8cos4theta$

$= \frac{5}{8} + \frac{3}{8} cos 4 θ$ $=5/8+3/8cos4theta$

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Question #02a6f

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