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

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

Question

(B) Given $f_k(x) = \frac{1}{k}(\cos^k x + \sin^k x)$.For $k=4$, $f_4(x) = \frac{1}{4}(\cos^4 x + \sin^4 x)$.Using $\cos^4 x + \sin^4 x = (\cos^2 x +

Vedclass · Accepted Answer

$-\frac{1}{12}$

Vedclass · Answer

(B) Given $f_k(x) = \frac{1}{k}(\cos^k x + \sin^k x)$.For $k=4$, $f_4(x) = \frac{1}{4}(\cos^4 x + \sin^4 x)$.Using $\cos^4 x + \sin^4 x = (\cos^2 x + \sin^2 x)^2 - 2\sin^2 x \cos^2 x = 1 - \frac{1}{2}\sin^2(2x)$, we get $f_4(x) = \frac{1}{4}(1 - \frac{1}{2}\sin^2(2x))$.For $k=6$, $f_6(x) = \frac{1}{6}(\cos^6 x + \sin^6 x)$.Using $\cos^6 x + \sin^6 x = (\cos^2 x + \sin^2 x)(\cos^4 x - \cos^2 x \sin^2 x + \sin^4 x) = 1 - 3\sin^2 x \cos^2 x = 1 - \frac{3}{4}\sin^2(2x)$, we get $f_6(x) = \frac{1}{6}(1 - \frac{3}{4}\sin^2(2x))$.Now, $f_6(x) - f_4(x) = \frac{1}{6} - \frac{1}{8}\sin^2(2x) - (\frac{1}{4} - \frac{1}{8}\sin^2(2x)) = \frac{1}{6} - \frac{1}{4} = \frac{2-3}{12} = -\frac{1}{12}$.

Let $f_k(x) = \frac{1}{k}(\cos^k x + \sin^k x)$ where $k \in \mathbb{N}$. Find the value of $f_6(x) - f_4(x)$.

Similar Questions

If $\sec \theta = m$ and $\tan \theta = n$,then $\frac{1}{m} \left[ m + n + \frac{1}{m + n} \right] = $

$\theta$ and $\alpha$ lie in $Q_3$. If $\cos (\theta-\alpha), \cos \theta, \cos (\theta+\alpha)$ are in harmonic progression,then $\cos \theta \sec \frac{\alpha}{2} = $

The exact value of $\cos^2 73^\circ + \cos^2 47^\circ + (\cos 73^\circ \cdot \cos 47^\circ)$ is

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

If $\sum_{r=1}^{50} \tan ^{-1} \frac{1}{2 r^2}=p$,then $\tan p$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module