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

In a right-angled triangle,the hypotenuse is $2 \sqrt{2}$ times the perpendicular drawn from the opposite vertex. Then the other acute angles of the triangle are

The value of $\tan 9^{\circ} - \tan 27^{\circ} - \tan 63^{\circ} + \tan 81^{\circ}$ is

If $\tan x = (\sin 45^{\circ})(\cos 45^{\circ}) + \sin 30^{\circ}$,then the value of $x$ is (in $^{\circ}$)

If $A, B, C$ are the angles of a triangle,then $\sin^2 A + \sin^2 B + \sin^2 C - 2\cos A \cos B \cos C = $

If $\pi < \alpha < \frac{3\pi}{2}$,then $\sqrt{\frac{1 - \cos \alpha}{1 + \cos \alpha}} + \sqrt{\frac{1 + \cos \alpha}{1 - \cos \alpha}} = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module