Question

Prove that sin⁶θ + cos⁶θ = 1 – 3 sin²θ cos²θ​

Answer 1

sin^6 θ + cos^6 θ = (sin^2 θ)^3 + (cos^2 θ)^3

Using identity, a^3 + b^3 = (a+b)(a^2 + b^2 - ab)

(sin^2 θ)^3 + (cos^2 θ)^3 = (sin^2 θ + cos^2 θ)(sin^4 θ + cos^4 θ - sin^2 θ.cos^2 θ)

Now sin^2 θ + cos^2 θ = 1

(sin^2 θ + cos^2 θ)(sin^4 θ + cos^4 θ - sin^2 θ.cos^2 θ) = (sin^2 θ)^2 + (cos^2 θ)^2 - sin^2 θ.cos^2 θ)

Using identity a^2 + b^2 = (a + b)^2 - 2ab

(sin^2 θ)^2 + (cos^2 θ)^2 - sin^2 θ.cos^2 θ) =

(sin^2 θ + cos^2 θ)^2 - 2sin^2 θ.cos^2 θ - sin^2 θ.cos^2 θ = 1 - 3.sin^2 θ.cos^2 θ = sin^6 θ + cos^6 θ

Hence, proved.

Answer 2

Step-by-step explanation:

sin⁶θ+cos⁶θ

=(sin²θ)³+(cos²θ)³

by using the identity,a³+b³=(a+b)(a²+b²-ab)

(sin²θ)³+(cos²θ)³

=(sin²θ+cos²θ)((sin²θ)²+(cos²θ)²-sin²θcos²θ)

by using the identity,a²+b²=(a+b)²-2ab

(sin²θ)²+(cos²θ)²-sin²θ+cos²θ

=(sin²θ+cos²θ)-2(sin²θcos²θ)-sin²θcos²θ

=1-2sin²θcos²θ-sin²θcos²θ

=1-3sin²θcos²θ=sin⁶θ+cos⁶θ

hence proved

hope it helps you