If the period of the function f(x) = sin 5x c | vedclass

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

Question

(D) Given the function $f(x) = \sin 5x \cos 3x$. Using the trigonometric identity $2 \sin A \cos B = \sin(A+B) + \sin(A-B)$, we can rewrite the functi

Vedclass · Accepted Answer

$-1$

Vedclass · Answer

(D) Given the function $f(x) = \sin 5x \cos 3x$. Using the trigonometric identity $2 \sin A \cos B = \sin(A+B) + \sin(A-B)$, we can rewrite the function as: $f(x) = \frac{1}{2} [\sin(5x+3x) + \sin(5x-3x)] = \frac{1}{2} (\sin 8x + \sin 2x)$. The period of $\sin 8x$ is $T_1 = \frac{2\pi}{8} = \frac{\pi}{4}$. The period of $\sin 2x$ is $T_2 = \frac{2\pi}{2} = \pi$. The period of the sum of two periodic functions is the Least Common Multiple ($L$.$C$.$M$.) of their individual periods. $\alpha = 	ext{L.C.M.}\left(\frac{\pi}{4}, \piight) = \frac{	ext{L.C.M.}(\pi, \pi)}{	ext{H.C.F.}(4, 1)} = \frac{\pi}{1} = \pi$. Thus, $\alpha = \pi$. Finally, $\cos \alpha = \cos \pi = -1$.

If the period of the function $f(x) = \sin 5x \cos 3x$ is $\alpha$, then $\cos \alpha =$

Similar Questions

If $f(x) = \sin^2\left(\frac{\pi}{8} + \frac{x}{2}\right) - \sin^2\left(\frac{\pi}{8} - \frac{x}{2}\right)$,then the period of $f$ is

The period of the function $f(x) = \cos^2(\sin x) + \sin^2(\cos x)$ is

The period of $f(x) = x - [x]$,if it is periodic,is

Let $f$ and $g$ be periodic functions with the periods $T_{1}$ and $T_{2}$ respectively. Then $f+g$ is

Let $f(x) = \sin x + \cos(ax)$ be a periodic function. Then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module