Question

prove that sin ^6 theta +cos^6 theta + 3 sin^2 theta cos^2 theta = 1

Answer 1

Step-by-step explanation:

sin

6

θ+cos

6

θ+3sin

2

θcos

2

θ

⇒LHS=(sin

2

θ)

3

+(cos

2

θ)

3

+3sin

2

θcos

2

θ

Using, [a

3

+b

3

=(a+b)

3

−3ab(a+b)]

⇒LHS=(sin

2

θ+cos

2

θ)

3

−3sin

2

θcos

2

θ(sin

2

θ+cos

2

θ)

3

+3sin

2

θcos

2

θ

⇒LHS=1−3sin

2

θcos

2

θ+3sin

2

θcos

2

θ=1=RHS