Question

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

Answer 1

Answer:

sin²θ + cos²θ = 1

Therefore (sin²θ + cos²θ)³ = 1

or

(sin²θ)³ + (cos²θ)³ + 3sin²θ cos²θ(sin²θ + cos²θ) = 1

or,

sin^6θ + cos^6θ+ 3sin²θ cos²θ = 1

HOPE IT HELPS YOU ✌

Answer 2

Answer:

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