Question

Prove that:- sin 6 theta + cos 6 theta = 1-3sin 2 theta.cos 2 theta

Answer 1

(sin4+cos4)2=1-3(sin+cos)2tete
1=1-3
1=-2
sin6+cos6≠1-3sin2.cos2

Answer 2

✸CORRECT QUESTION:-

Prove that

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

✸PROOF:-

⠀⠀⠀⠀

We have LHS

$\implies \sf {sin}^{6} \theta + {cos}^{6} \theta$

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

✶Using below identity we have:-

$\sf \longrightarrow {a }^{3} + {b}^{3} = (a + b)({a}^{2} - ab + {b}^{2} )$

⠀⠀⠀⠀

$\implies \sf{ ( {sin}^{2} \theta + {cos}^{2} \theta)} \{(sin^{2} \theta)^{2} - sin^{2} \theta.cos^{2} \theta +({ cos}^{2} \theta)^{2} \}$

✶Using below identity we have:-⠀⠀⠀

$\sf \longrightarrow sin^{2} \theta + cos^{2} \theta$

⠀⠀⠀⠀

$\implies \sf(1)\{(sin^{2} \theta)^{2} - sin^{2} \theta.cos^{2} \theta +({ cos}^{2} \theta)^{2} \}$

$\sf \implies \{sin^{4} \theta + cos^{4} \theta - sin^{2} \theta.cos^{ 2} \theta \}$

⠀⠀⠀⠀

✶Using below algebraic identity we have:-

$\longrightarrow \sf a^{2} + b ^{2} = {(a + b)}^{2} - 2ab$

⠀⠀⠀⠀

$\sf \implies \{( {sin}^{2} \theta + cos^{2} \theta)^{2} - 2sin^{2} \theta.cos ^{2} \theta - sin^{2} \theta.cos^{2} \theta \}$

$\sf \implies(1 - 3sin^{2} \theta.cos^{2} \theta)$

Therefore LHS=RHS

Hence ,proved