Question

if sin theta + cos theta equal to X prove that sin to the power 6 theta + cos to the power 6 theta = 4-3(x^2-1)^2/4

Answer 1

Answer is in the attachment.
Hope it helped u .

Answer 2

Answer: Proved.

Step-by-step explanation: Given that

$\sin\theta+\cos\theta=x.$

We are given to prove the following:

$\sin^6\theta+\cos^6\theta=\dfrac{4-3(x^2-1)^2}{4}.$

We have

$\sin\theta+\cos\theta=x\\\\\Rightarrow (\sin\theta+\cos\theta)^2=x^2\\\\\Rightarrow \sin^2\theta+\cos^2\theta+2\sin\theta\cos\theta=x^2\\\\\Rightarrow 1+2\sin\theta\cos\theta=x^2\\\\\Rightarrow \sin\theta\cos\theta=\dfrac{x^2-1}{2}.~~~~~~~~~~~~~~~~~~(i)$

Therefore,

$L.H.S.\\\\=\cos^6\theta+\sin^6\theta\\\\=(\cos^2\theta)^3+(\sin^2\theta)^3\\\\=(\cos^2\theta+\sin^2\theta)(\cos^4\theta+\sin^4\theta-\cos^2\theta\sin^2\theta)\\\\=1\times \left((\cos^2\theta+\sin^2\theta)^2-3\cos^2\theta\sin^2\theta\right)\\\\=1-3\cos^2\theta\sin^2\theta\\\\=1-3\left(\dfrac{x^2-1}{2}\right)^2~~~~~~~~~\textup{(Using equation (i))}\\\\\\=1-\dfrac{3}{4}(x^2-1)^2\\\\\\=\dfrac{4-3(x^2-1)^2}{4}\\\\=R.H.S.$

Hence proved.