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

If $A = \sin^2 x + \cos^4 x$,then for all real $x :$

If $\frac{\sin(x + y)}{\sin(x - y)} = \frac{a + b}{a - b},$ then $\frac{\tan x}{\tan y}$ is equal to

$\cos^2 A(3 - 4\cos^2 A)^2 + \sin^2 A(3 - 4\sin^2 A)^2 = $

If $\theta+\phi=\frac{2 \pi}{3}$ and $\cos \theta=\frac{\sqrt{3}}{2},$ what is the value of $\sin \phi ?$

If $\sin x + \sin y = 3(\cos y - \cos x),$ then the value of $\frac{\sin 3x}{\sin 3y}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module