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) We are 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

Vedclass · Accepted Answer

$\frac{1}{12}$

Vedclass · Answer

(B) We are given $f_k(x) = \frac{1}{k}(\sin^k x + \cos^k x)$.We need to find $f_4(x) - f_6(x)$.$f_4(x) = \frac{1}{4}(\sin^4 x + \cos^4 x) = \frac{1}{4}((\sin^2 x + \cos^2 x)^2 - 2\sin^2 x \cos^2 x) = \frac{1}{4}(1 - 2\sin^2 x \cos^2 x)$.$f_6(x) = \frac{1}{6}(\sin^6 x + \cos^6 x) = \frac{1}{6}((\sin^2 x + \cos^2 x)(\sin^4 x - \sin^2 x \cos^2 x + \cos^4 x)) = \frac{1}{6}(1 - 3\sin^2 x \cos^2 x)$.Now,$f_4(x) - f_6(x) = \frac{1}{4}(1 - 2\sin^2 x \cos^2 x) - \frac{1}{6}(1 - 3\sin^2 x \cos^2 x)$.$= (\frac{1}{4} - \frac{1}{6}) - (\frac{2}{4} - \frac{3}{6})\sin^2 x \cos^2 x$.$= (\frac{3-2}{12}) - (\frac{1}{2} - \frac{1}{2})\sin^2 x \cos^2 x$.$= \frac{1}{12} - 0 = \frac{1}{12}$.

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 $\cos x - \sin x = \sqrt{a} \sin x$,then $a \sin x + \cos x - \sin x = $

$\sin^6 \theta + \cos^6 \theta + 3 \sin^2 \theta \cos^2 \theta = $

If $\cos \theta - 4 \sin \theta = 1$,then $\sin \theta + 4 \cos \theta$ is equal to

The value of $\cos^2 10^o - \cos 10^o \cos 50^o + \cos^2 50^o$ is

$16 \sin(20^{\circ}) \sin(40^{\circ}) \sin(80^{\circ})$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module