Let alpha, beta be such that pi

Question

(D) Given $\sin \alpha + \sin \beta = -\frac{21}{65}$ and $\cos \alpha + \cos \beta = -\frac{27}{65}$.Squaring and adding both equations:$(\sin \alpha

Vedclass · Accepted Answer

$-\frac{3}{\sqrt{130}}$

Vedclass · Answer

(D) Given $\sin \alpha + \sin \beta = -\frac{21}{65}$ and $\cos \alpha + \cos \beta = -\frac{27}{65}$.Squaring and adding both equations:$(\sin \alpha + \sin \beta)^2 + (\cos \alpha + \cos \beta)^2 = \left(-\frac{21}{65}\right)^2 + \left(-\frac{27}{65}\right)^2$$(\sin^2 \alpha + \cos^2 \alpha) + (\sin^2 \beta + \cos^2 \beta) + 2(\sin \alpha \sin \beta + \cos \alpha \cos \beta) = \frac{441 + 729}{4225}$$1 + 1 + 2\cos(\alpha - \beta) = \frac{1170}{4225}$$2 + 2\cos(\alpha - \beta) = \frac{1170}{4225} = \frac{18}{65}$$2(1 + \cos(\alpha - \beta)) = \frac{18}{65}$$4\cos^2\left(\frac{\alpha - \beta}{2}\right) = \frac{18}{65}$$\cos^2\left(\frac{\alpha - \beta}{2}\right) = \frac{18}{260} = \frac{9}{130}$$\cos\left(\frac{\alpha - \beta}{2}\right) = \pm \frac{3}{\sqrt{130}}$Since $\pi In this interval,the cosine function is negative.Therefore,$\cos\left(\frac{\alpha - \beta}{2}\right) = -\frac{3}{\sqrt{130}}$.

Let $\alpha, \beta$ be such that $\pi < (\alpha - \beta) < 3\pi$. If $\sin \alpha + \sin \beta = -\frac{21}{65}$ and $\cos \alpha + \cos \beta = -\frac{27}{65}$,then the value of $\cos \frac{\alpha - \beta}{2}$ is

Similar Questions

$\frac{1 + \sin A - \cos A}{1 + \sin A + \cos A} = $

If $\tan A - \tan B = x$ and $\cot B - \cot A = y,$ then $\cot (A - B) = $

If $\sin A + \cos A = \sqrt{2}$,then $\cos^2 A = $

If $\sin \theta + \sin \phi = a$ and $\cos \theta + \cos \phi = b,$ then $\tan \frac{\theta - \phi}{2}$ is equal to

In any triangle $ABC$,the expression ${\sin ^2}\frac{A}{2} + {\sin ^2}\frac{B}{2} + {\sin ^2}\frac{C}{2}$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module