Question

2 ( sin⁶ theta +cos⁶ theta) - 3(sin⁴ theta + cos ⁴ theta +1 )+1​

Answer 1

ANSWER:

To Solve:

• 2 (sin⁶ theta + cos⁶ theta) - 3(sin⁴ theta + cos⁴ theta + 1) + 1

Solution:

We are given that,

$\implies2(\sin^6\theta+\cos^6\theta)-3(\sin^4\theta+\cos^4\theta+1)+1$

We can rewrite it as,

$\implies2\bigg((\sin^2\theta)^3+(\cos^2\theta)^3\bigg)-3\bigg((\sin^2\theta)^2+(\cos^2\theta)^2+1\bigg)+1$

We know that,

$\hookrightarrow a^3+b^3=(a+b)^3-3(ab)(a+b)$

And,

$\hookrightarrow a^2+b^2=(a+b)^2-2ab$

So,

$\implies2\bigg((\sin^2\theta)^3+(\cos^2\theta)^3\bigg)-3\bigg((\sin^2\theta)^2+(\cos^2\theta)^2+1\bigg)+1$

$\implies2\bigg((\sin^2\theta+\cos^2\theta)^3-3(\sin^2\theta\cos^2\theta)(\sin^2\theta+\cos^2\theta)\bigg)-3\bigg((\sin^2\theta+\cos^2\theta)^2-2(\sin^2\theta\cos^2\theta)+1\bigg)+1$

We know that,

$\hookrightarrow \sin^2\phi+\cos^2\phi=1$

So,

$\implies2\bigg((\sin^2\theta+\cos^2\theta)^3-3(\sin^2\theta\cos^2\theta)(\sin^2\theta+\cos^2\theta)\bigg)-3\bigg((\sin^2\theta+\cos^2\theta)^2-2(\sin^2\theta\cos^2\theta)+1\bigg)+1$

$\implies2\bigg((1)^3-3(\sin^2\theta\cos^2\theta)(1)\bigg)-3\bigg((1)^2-2(\sin^2\theta\cos^2\theta)+1\bigg)+1$

$\implies2\bigg(1-3\sin^2\theta\cos^2\theta\bigg)-3\bigg(1-2\sin^2\theta\cos^2\theta+1\bigg)+1$

$\implies2\bigg(1-3\sin^2\theta\cos^2\theta\bigg)-3\bigg(2-2\sin^2\theta\cos^2\theta\bigg)+1$

$\implies2-6\sin^2\theta\cos^2\theta-6+6\sin^2\theta\cos^2\theta+1$

Cancelling, 6sin^2 theta cos^2 theta,

$\implies2-6+1$

So,

$\implies -3$

Hence,

$\implies\bf2(sin^6\theta+cos^6\theta)-3(sin^4\theta+cos^4\theta+1)+1=-3$