Question

sin^6theta+cos^6theta=1-3 sin^2 theta cos^2 theta​

Answer 1

Qᴜᴇsᴛɪᴏɴ :-

Prove :- sin⁶θ + cos⁶θ = 1 - 3 sin²θ*cos²θ

Sᴏʟᴜᴛɪᴏɴ :-

Taking LHS ,

→ sin⁶θ + cos⁶θ

Now a⁶ can be written as (a²)³ ,

→ (sin²θ)³ + (cos²θ)³

Now, using a³ + b³ = (a + b)³ - 3ab(a + b)

→ (sin²θ + cos²θ)³ - 3 * sin²θ*cos²θ * (sin²θ + cos²θ)

Now using sin²A + cos²A = 1,

→ 1 - 3 * sin²θ*cos²θ * 1

→ 1 - 3sin²θ*cos²θ = RHS (Proved).

Answer 2

${\huge{\bf{\red{\underline{Solution:}}}}}$

${\bf{\blue{\underline{Take\:L.H.S:}}}}$

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

$: \implies{\sf{ { \big [ \: (sin \theta}^{2} )^{3} + ( {cos}^{2}) ^{3} \big] }}$

We know that,

$: \boxed{\sf{ \purple{ {x}^{3} + {y}^{3} = (x + y)^{3} - 3xy(x + y) }}}$

$: \implies{\sf{ \big[ ( {sin}^{2} \theta + {cos}^{2} \theta )\big] ^{3} - 3 {sin}^{2} \theta {cos}^{2} \theta( {sin}^{2} \theta + {cos}^{2} \theta) }}$

$: \boxed{\sf{ \purple{ {sin}^{2} \theta + {cos}^{2} \theta = 1 }}}$

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

$: \implies{\sf{ 1 - 3 {sin}^{2} \theta {cos}^{2} \theta }}$

Hence L.HS=R.H.S