Question

$cos^6theta - sin^6theta= costheta(1-1/4sin^2 2theta)​$
Solve The Math..​

Answer 1

Explanation:

L.H.S. =sin

6

θ+cos

6

θ

=(sin

2

θ)

3

+(cos

2

θ)

3

Put sin

2

θ=a and cos

2

θ=b

∴ L.H.S. =a

3

+b

3

=(a+b)

3

−2ab(a+b)

=(sin

2

θ−cos

2

θ)

3

−3sin

2

θcos

2

θ(sin

2

θ+cos

2

θ)

=(1)

3

−3sin

2

θcos

2

θ[∵sin

2

θ+cos

2

θ=1]

=1−3sin

2

θcos

2

θ

= R.H.S.

Answer 2

Explanation:

Dear student.

(cos2a)3 – (sin2a)3

= (cos2a – sin2a )(cos4a + sin4a + cos2a*sin2a)

= (cos2a)(1 – 2 cos2a*sin2a + cos2a*sin2a)

= (cos2a)(1 – cos2a*sin2a )

= (cos2a)(1 – 4cos2a*sin2a/4 )

= cos2a(1-1/4sin22a)