Question

evaluate: sin^6 alpha+cos^6alpha=1-3sin^2alpha+3sin^4alpha​

Answer 1

Answer:

to find the value of alpha???????

Answer 2

$\sin^6 \alpha+\cos^6\alpha=1-3\sin^2\alpha+3\sin^4\alpha,$ proved.

Step-by-step explanation:

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

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

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

Using the algebraic identity,

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

⇒ $a^{3} +b^{3}=(a+b)^{3}-3ab(a+b)$

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

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

Using the trigonometric identity,

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

$=1-3\sin^2 \alpha\cos^2\alpha$

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

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

= R.H.S., proved.

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