Question

prove that sin^6alpha + cos^6alpha= 3sin⁴alpha + 3cos²alpha - 2​

Answer 1

Answer:

prove that, \sin^6 \alpha+\cos^6\alpha=1-3\sin^2\alpha+3\sin^4\alpha.sin

6

α+cos

6

α=1−3sin

2

α+3sin

4

α.

L.H.S. =\sin^6 \alpha+\cos^6\alpha=sin

6

α+cos

6

α

=(\sin^2 \alpha)^3+(\cos^2\alpha)^3=(sin

2

α)

3

+(cos

2

α)

3

Using the algebraic identity,

(a+b)^{3} =a^{3} +b^{3}+3ab(a+b)(a+b)

3

=a

3

+b

3

+3ab(a+b)

⇒ a^{3} +b^{3}=(a+b)^{3}-3ab(a+b)a

3

+b

3

=(a+b)

3

−3ab(a+b)

=(\sin^2 \alpha+\cos^2\alpha)^3-3\sin^2 \alpha\cos^2\alpha(\sin^2 \alpha+\cos^2\alpha)=(sin

2

α+cos

2

α)

3

−3sin

2

αcos

2

α(sin

2

α+cos

2

α)

=(1)^3-3\sin^2 \alpha\cos^2\alpha(1)=(1)

3

−3sin

2

αcos

2

α(1)

Using the trigonometric identity,

\sin^2 A+\cos^2A=1sin

2

A+cos

2

A=1

=1-3\sin^2 \alpha\cos^2\alpha=1−3sin

2

αcos

2

α

=1-3\sin^2 \alpha(1-\sin^2 \alpha)=1−3sin

2

α(1−sin

2

α)

=1-3\sin^2 \alpha+3\sin^4 \alpha=1−3sin

2

α+3sin

4

α

= R.H.S., proved.

Thus, \sin^6 \alpha+\cos^6\alpha=1-3\sin^2\alpha+3\sin^4\alpha,sin

6

α+cos

6

α=1−3sin

2

α+3sin

4

α, proved.