If cos alpha + cos beta = frac{3}{2} and sin | vedclass

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

Question

(B) Given $\cos \alpha + \cos \beta = \frac{3}{2}$ and $\sin \alpha + \sin \beta = \frac{1}{2}$.Using sum-to-product formulas:$2 \cos \frac{\alpha+\be

Vedclass · Accepted Answer

$\frac{7}{5}$

Vedclass · Answer

(B) Given $\cos \alpha + \cos \beta = \frac{3}{2}$ and $\sin \alpha + \sin \beta = \frac{1}{2}$.Using sum-to-product formulas:$2 \cos \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2} = \frac{3}{2}$ $(i)$$2 \sin \frac{\alpha+\beta}{2} \cos \frac{\alpha-\beta}{2} = \frac{1}{2}$ $(ii)$Dividing $(ii)$ by $(i)$ gives $	an \frac{\alpha+\beta}{2} = \frac{1}{3}$.Since $	heta = \frac{\alpha+\beta}{2}$,we have $	an 	heta = \frac{1}{3}$.We need to find $\sin 2	heta + \cos 2	heta = \frac{2 	an 	heta}{1 + 	an^2 	heta} + \frac{1 - 	an^2 	heta}{1 + 	an^2 	heta}$.Substituting $	an 	heta = \frac{1}{3}$:$= \frac{2(1/3)}{1 + 1/9} + \frac{1 - 1/9}{1 + 1/9} = \frac{2/3}{10/9} + \frac{8/9}{10/9} = \frac{6}{10} + \frac{8}{10} = \frac{14}{10} = \frac{7}{5}$.

If $\cos \alpha + \cos \beta = \frac{3}{2}$ and $\sin \alpha + \sin \beta = \frac{1}{2}$ and $\theta$ is the arithmetic mean of $\alpha$ and $\beta$,then $\sin 2\theta + \cos 2\theta$ is equal to

Similar Questions

Evaluate: $\sin ^2 18^{\circ}+\sin ^2 24^{\circ}+\sin ^2 36^{\circ}+\sin ^2 42^{\circ}+\sin ^2 78^{\circ}+\sin ^2 90^{\circ}+\sin ^2 96^{\circ}+\sin ^2 102^{\circ}+\sin ^2 138^{\circ}+\sin ^2 162^{\circ}$

Let $f_k(x) = \frac{1}{k}(\cos^k x + \sin^k x)$ where $k \in \mathbb{N}$. Find the value of $f_6(x) - f_4(x)$.

Let $S$ be the set of all $\alpha \in \mathbb{R}$ such that the equation $\cos 2x + \alpha \sin x = 2\alpha - 7$ has a solution. Then $S$ is equal to

If $a = \sin \frac{\pi}{18} \sin \frac{5\pi}{18} \sin \frac{7\pi}{18}$ and $x$ is the solution of the equations $y = 2[x] + 2$ and $y = 3[x - 2]$,where $[x]$ denotes the greatest integer function of $x$,then $a$ is equal to:

If $\sin A+\sin B=\sqrt{3}(\cos B-\cos A)$,then $\sin 3A+\sin 3B$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module