Question

10. Prove that: sin6θ + cos6θ + 3sin2θ cos2θ = 1.

Answer 1

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FORMULA TO BE IMPLEMENTED

1.

${sin}^{2} \theta + {cos}^{2} \theta = 1$

2.

${(x + y)}^{3} = {x}^{3} + {y}^{3} + 3xy(x + y)$

TO PROVE

${sin}^{6} \theta + {cos}^{6} \theta + 3 {sin}^{2} \theta {cos}^{2} \theta = 1$

CALCULATION

${sin}^{6} \theta + {cos}^{6} \theta + 3 {sin}^{2} \theta {cos}^{2} \theta$

$= {({sin}^{2} \theta )}^{3} + {( {cos}^{2} \theta) }^{3} + 3 {sin}^{2} \theta {cos}^{2} \theta$

$= {({sin}^{2} \theta )}^{3} + {( {cos}^{2} \theta) }^{3} + 3 \times {sin}^{2} \theta \times {cos}^{2} \theta \: ({sin}^{2} \theta + {cos}^{2} \theta )$

$= {( \: {sin}^{2} \theta + {cos}^{2} \theta \:) }^{3}$

$= { \: (1)}^{3}$

$= 1$

$\fbox{HENCE PROVED }$