Question

prove that
sin6theta +cos6theta + 3sin2theta*cos2theta=1​

Answer 1

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Answer 2

Answer:

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