sin left(frac{pi}{3}+xright)-cos left(frac{pi | vedclass

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

Question

(D) We use the expansion formulas: $\sin(A+B) = \sin A \cos B + \cos A \sin B$ and $\cos(A+B) = \cos A \cos B - \sin A \sin B$. $\sin \left(\frac{\pi}

Vedclass · Accepted Answer

$\sin x$

Vedclass · Answer

(D) We use the expansion formulas: $\sin(A+B) = \sin A \cos B + \cos A \sin B$ and $\cos(A+B) = \cos A \cos B - \sin A \sin B$. $\sin \left(\frac{\pi}{3}+x\right) = \sin \frac{\pi}{3} \cos x + \cos \frac{\pi}{3} \sin x = \frac{\sqrt{3}}{2} \cos x + \frac{1}{2} \sin x$. $\cos \left(\frac{\pi}{6}+x\right) = \cos \frac{\pi}{6} \cos x - \sin \frac{\pi}{6} \sin x = \frac{\sqrt{3}}{2} \cos x - \frac{1}{2} \sin x$. Subtracting the two expressions: $\left(\frac{\sqrt{3}}{2} \cos x + \frac{1}{2} \sin x\right) - \left(\frac{\sqrt{3}}{2} \cos x - \frac{1}{2} \sin x\right)$ $= \frac{\sqrt{3}}{2} \cos x + \frac{1}{2} \sin x - \frac{\sqrt{3}}{2} \cos x + \frac{1}{2} \sin x$ $= \sin x$.

$\sin \left(\frac{\pi}{3}+x\right)-\cos \left(\frac{\pi}{6}+x\right) = $

Similar Questions

If $\sin (A+B) \sin (A-B)+\cos (A+B) \cos (A-B)=\frac{1}{2}$ and $0 < B < \frac{\pi}{2}$,then $B=$

If $\cos \alpha + \cos \beta = \frac{1}{3}$ and $\sin \alpha + \sin \beta = \frac{1}{4}$,then $\cos (\alpha + \beta) = $

If $\sin A = \frac{4}{5}$ and $\cos B = -\frac{12}{13}$,where $A$ and $B$ lie in the first and third quadrant respectively,then $\cos(A + B) = $

If $\cos \theta = \frac{3}{5}$ and $\cos \phi = \frac{4}{5},$ where $\theta$ and $\phi$ are positive acute angles,then $\cos \frac{\theta - \phi}{2} = $

If $\theta+\phi=\frac{\pi}{4}$,then $(1+\tan \theta)(1+\tan \phi)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module