Let f_k(x) = frac{1}{k}(sin^k x + cos^k x) wh | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) Given $f_k(x) = \frac{1}{k}(\sin^k x + \cos^k x)$.We need to find $f_4(x) - f_6(x)$.$f_4(x) = \frac{1}{4}(\sin^4 x + \cos^4 x) = \frac{1}{4}((\sin

Vedclass · Accepted Answer

$\frac{1}{12}$

Vedclass · Answer

(B) Given $f_k(x) = \frac{1}{k}(\sin^k x + \cos^k x)$.We need to find $f_4(x) - f_6(x)$.$f_4(x) = \frac{1}{4}(\sin^4 x + \cos^4 x) = \frac{1}{4}((\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x) = \frac{1}{4}(1 - 2\sin^2 x \cos^2 x)$.$f_6(x) = \frac{1}{6}(\sin^6 x + \cos^6 x) = \frac{1}{6}((\sin^2 x + \cos^2 x)(\sin^4 x + \cos^4 x - \sin^2 x \cos^2 x)) = \frac{1}{6}(1 - 3\sin^2 x \cos^2 x)$.Now,$f_4(x) - f_6(x) = \frac{1}{4}(1 - 2\sin^2 x \cos^2 x) - \frac{1}{6}(1 - 3\sin^2 x \cos^2 x)$.$= (\frac{1}{4} - \frac{1}{2}\sin^2 x \cos^2 x) - (\frac{1}{6} - \frac{1}{2}\sin^2 x \cos^2 x)$.$= \frac{1}{4} - \frac{1}{6} = \frac{3 - 2}{12} = \frac{1}{12}$.

Let $f_k(x) = \frac{1}{k}(\sin^k x + \cos^k x)$ where $x \in R$ and $k \ge 1$. Then $f_4(x) - f_6(x)$ is equal to:

Similar Questions

If $\tan \beta = \cos \theta \tan \alpha ,$ then ${\tan ^2}\frac{\theta }{2} = $

$\cos 2\theta + 2\cos \theta$ is always

$(\sec A + \tan A - 1)(\sec A - \tan A + 1) - 2\tan A = $

The expression $\frac{\tan A}{1 - \cot A} + \frac{\cot A}{1 - \tan A}$ can be written as:

$\frac{\sin^2 A - \sin^2 B}{\sin A \cos A - \sin B \cos B} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module