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

Two straight roads intersect at an angle of $60^o$. $A$ bus on one road is $2 \, km$ away from the intersection and a car on the other road is $3 \, km$ away from the intersection. Then the direct distance between the two vehicles is

If $0 < \theta < \frac{\pi}{2}$ and $\tan 3 \theta \neq 0$, then $\tan \theta + \tan 2 \theta + \tan 3 \theta = 0$ if $\tan \theta \cdot \tan 2 \theta = k$, where $k =$

$\operatorname{cosec} 18^{\circ}$ is a root of the equation :

The smallest positive value (in degrees) of $\theta$ for which $\tan(\theta+100^{\circ})=\tan(\theta+50^{\circ}) \tan(\theta) \tan(\theta-50^{\circ})$ is valid,is (in $^{\circ}$)

If $\operatorname{cosec}^2(\alpha+\beta)-\sin^2(\beta-\alpha)+\sin^2(2\alpha-\beta)=\cos^2(\alpha-\beta)$ where $\alpha, \beta \in (0, \frac{\pi}{2})$,then $\sin(\alpha-\beta)$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module