If A+B+C+D=2 pi,then cos A-cos B+cos C-cos D= | vedclass

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

Question

(C) Given $A+B+C+D=2 \pi$. We have $\cos A - \cos B + \cos C - \cos D = (\cos A - \cos B) + (\cos C - \cos D)$. Using the formula $\cos X - \cos Y = -

Vedclass · Accepted Answer

$-4 \sin \frac{A+B}{2} \sin \frac{A+C}{2} \sin \frac{A+D}{2}$

Vedclass · Answer

(C) Given $A+B+C+D=2 \pi$. We have $\cos A - \cos B + \cos C - \cos D = (\cos A - \cos B) + (\cos C - \cos D)$. Using the formula $\cos X - \cos Y = -2 \sin \frac{X+Y}{2} \sin \frac{X-Y}{2}$,we get: $= -2 \sin \frac{A+B}{2} \sin \frac{A-B}{2} - 2 \sin \frac{C+D}{2} \sin \frac{C-D}{2}$. Since $A+B+C+D=2 \pi$,we have $\frac{C+D}{2} = \pi - \frac{A+B}{2}$,so $\sin \frac{C+D}{2} = \sin \frac{A+B}{2}$. Also,$\frac{C-D}{2} = \frac{(2 \pi - A - B) - 2D}{2} = \pi - \frac{A+B+2D}{2}$. Using the identity,the expression simplifies to: $-4 \sin \frac{A+B}{2} \sin \frac{A+C}{2} \sin \frac{A+D}{2}$.

If $A+B+C+D=2 \pi$,then $\cos A-\cos B+\cos C-\cos D=$

Similar Questions

$\cos 13^{\circ} \sin 17^{\circ} \sin 21^{\circ} \cos 47^{\circ} = $

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

If $\sin \alpha = \sin \beta$ and $\cos \alpha = \cos \beta$,then $\alpha - \beta = $ for some integer $n$.

$\cos^2 76^{\circ} + \cos^2 16^{\circ} - \cos 76^{\circ} \cos 16^{\circ}$ is equal to

The product $(1+\tan 1^{\circ})(1+\tan 2^{\circ})(1+\tan 3^{\circ}) \dots (1+\tan 45^{\circ})$ equals

Vedclass Test Series

Exam Paper Generator

Online Exam Module