Question

Prove that 3(sin theta -cos theta)^4+6(sin theta+cos theta)^2 +4(sin^6theta+cos^6theta) is independent of theta.

Answer 1

we have to prove that, $3(sin\theta-cos\theta)^4+6(sin\theta+cos\theta)^2+4(sin^6\theta+cos^6\theta)$ is independent of $\theta$.

first of all, resolve the $3(sin\theta-cos\theta)^4+6(sin\theta+cos\theta)^2+4(sin^6\theta+cos^6\theta)$

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

we know, sin²x + cos²x = 1

so, $sin^2\theta+cos^2\theta=1$

and also use the formula, $2sin\theta cos\theta=sin2\theta$

= $3(1-sin2\theta)^2+6(1+sin2\theta)+4(sin^4\theta+cos^4\theta-sin^2\theta cos^2\theta)$

= $3(1+sin^22\theta-2sin2\theta)+6+6sin2\theta+4\{(sin^2\theta+cos^2\theta)^2-2sin^2\theta cos^2\theta-sin^2\theta cos^2\theta\}$

= $3+3sin^22\theta-6sin2\theta+6+6sin2\theta+4\{1-3/4sin^22\theta\}$

= $9+4+3sin^22\theta-3sin^22\theta$

= 13

here we see that, resolution of given trigonometric expression doesn't involve $\theta$ hence, it is independent of $\theta$

Answer 2

Step-by-step explanation: please go through the photo attached and please mark brainliest