If A+B+C=270^{circ},then cos 2A + cos 2B + co | vedclass

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

Question

(C) Given that $A+B+C=270^{\circ}$.We need to evaluate $\cos 2A + \cos 2B + \cos 2C + 4 \sin A \sin B \sin C$.Using the sum-to-product formula $\cos 2

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(C) Given that $A+B+C=270^{\circ}$.We need to evaluate $\cos 2A + \cos 2B + \cos 2C + 4 \sin A \sin B \sin C$.Using the sum-to-product formula $\cos 2B + \cos 2C = 2 \cos(B+C) \cos(B-C)$,we get:$= \cos 2A + 2 \cos(B+C) \cos(B-C) + 4 \sin A \sin B \sin C$.Since $B+C = 270^{\circ} - A$,then $\cos(B+C) = \cos(270^{\circ} - A) = -\sin A$.Substituting this:$= \cos 2A + 2(-\sin A) \cos(B-C) + 4 \sin A \sin B \sin C$$= (1 - 2 \sin^2 A) - 2 \sin A \cos(B-C) + 4 \sin A \sin B \sin C$$= 1 - 2 \sin A [\sin A + \cos(B-C)] + 4 \sin A \sin B \sin C$.Since $\sin A = \sin(270^{\circ} - (B+C)) = -\cos(B+C)$:$= 1 - 2 \sin A [-\cos(B+C) + \cos(B-C)] + 4 \sin A \sin B \sin C$.Using the identity $-\cos(B+C) + \cos(B-C) = 2 \sin B \sin C$:$= 1 - 2 \sin A (2 \sin B \sin C) + 4 \sin A \sin B \sin C$$= 1 - 4 \sin A \sin B \sin C + 4 \sin A \sin B \sin C$$= 1$.

If $A+B+C=270^{\circ}$,then $\cos 2A + \cos 2B + \cos 2C + 4 \sin A \sin B \sin C =$

Similar Questions

The minimum value of $\frac{\tan(x + \frac{\pi}{6})}{\tan x}$ is

The minimum value of $3 \cos x + 4 \sin x + 5$ is

In a triangle $ABC,$ the value of $\sin A + \sin B + \sin C$ is

If $A, B, C$ are the angles of a triangle,then $\sin 2A - \sin 2B + \sin 2C =$

If $\alpha+\beta=\frac{\pi}{2}$ and $\beta+\gamma=\alpha$,then $\tan \alpha$ equals

Vedclass Test Series

Exam Paper Generator

Online Exam Module