If sin x + sin y = frac{sqrt{3}+1}{2} and cos | vedclass

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

Question

(A) Given: $\sin x + \sin y = \frac{\sqrt{3}+1}{2}$ and $\cos x + \cos y = \frac{\sqrt{3}-1}{2}$.Using sum-to-product formulas:$2 \sin \left(\frac{x+y

Vedclass · Accepted Answer

$8+4 \sqrt{3}$

Vedclass · Answer

(A) Given: $\sin x + \sin y = \frac{\sqrt{3}+1}{2}$ and $\cos x + \cos y = \frac{\sqrt{3}-1}{2}$.Using sum-to-product formulas:$2 \sin \left(\frac{x+y}{2}ight) \cos \left(\frac{x-y}{2}ight) = \frac{\sqrt{3}+1}{2} \implies \sin \left(\frac{x+y}{2}ight) \cos \left(\frac{x-y}{2}ight) = \frac{\sqrt{3}+1}{4} \quad (i)$$2 \cos \left(\frac{x+y}{2}ight) \cos \left(\frac{x-y}{2}ight) = \frac{\sqrt{3}-1}{2} \implies \cos \left(\frac{x+y}{2}ight) \cos \left(\frac{x-y}{2}ight) = \frac{\sqrt{3}-1}{4} \quad (ii)$Dividing $(i)$ by $(ii)$:$	an \left(\frac{x+y}{2}ight) = \frac{\sqrt{3}+1}{\sqrt{3}-1} = \frac{(\sqrt{3}+1)^2}{3-1} = \frac{4+2\sqrt{3}}{2} = 2+\sqrt{3}$.So,$	an^2 \left(\frac{x+y}{2}ight) = (2+\sqrt{3})^2 = 4+3+4\sqrt{3} = 7+4\sqrt{3}$.Squaring and adding $(i)$ and $(ii)$:$\cos^2 \left(\frac{x-y}{2}ight) \left[ \sin^2 \left(\frac{x+y}{2}ight) + \cos^2 \left(\frac{x+y}{2}ight) ight] = \left(\frac{\sqrt{3}+1}{4}ight)^2 + \left(\frac{\sqrt{3}-1}{4}ight)^2$$\cos^2 \left(\frac{x-y}{2}ight) = \frac{3+1+2\sqrt{3} + 3+1-2\sqrt{3}}{16} = \frac{8}{16} = \frac{1}{2}$.$\sec^2 \left(\frac{x-y}{2}ight) = 2 \implies 	an^2 \left(\frac{x-y}{2}ight) = 2-1 = 1$.Therefore,$	an^2 \left(\frac{x-y}{2}ight) + 	an^2 \left(\frac{x+y}{2}ight) = 1 + 7 + 4\sqrt{3} = 8+4\sqrt{3}$.

If $\sin x + \sin y = \frac{\sqrt{3}+1}{2}$ and $\cos x + \cos y = \frac{\sqrt{3}-1}{2}$,then $\tan^2 \left(\frac{x-y}{2}\right) + \tan^2 \left(\frac{x+y}{2}\right) = $

Similar Questions

The number of ordered pairs $(x, y)$ satisfying the equations $\sin x + \sin y = \sin(x + y)$ and $|x| + |y| = 1$ is

The value of $\frac{\sin^2 \theta}{\sin \theta - \cos \theta} - \frac{\sin \theta + \cos \theta}{\tan^2 \theta - 1}$ for all permissible values of $\theta$ is:

The value of $\left( 1 + \cos \frac{\pi }{9} \right) \left( 1 + \cos \frac{3\pi }{9} \right) \left( 1 + \cos \frac{5\pi }{9} \right) \left( 1 + \cos \frac{7\pi }{9} \right)$ is

If $\tan \left(\frac{\pi}{9}\right), x, \tan \left(\frac{7 \pi}{18}\right)$ are in arithmetic progression and $\tan \left(\frac{\pi}{9}\right), y, \tan \left(\frac{5 \pi}{18}\right)$ are also in arithmetic progression,then $|x-2 y|$ is equal to:

If $\sec(\theta+\alpha)$,$\sec\theta$,and $\sec(\theta-\alpha)$ are in arithmetic progression,then the value of $\cos\theta \cdot \sec\frac{\alpha}{2}$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module