If A+B+C=2S,then sin(2S-A)+sin(2S-B)+sin(2S-C | vedclass

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

Question

(B) Given $A+B+C = 2S$. Then $2S-A = B+C$,$2S-B = A+C$,and $2S-C = A+B$. The expression becomes $\sin(B+C) + \sin(A+C) + \sin(A+B) - \sin(2S)$. Using

Vedclass · Accepted Answer

$4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$

Vedclass · Answer

(B) Given $A+B+C = 2S$. Then $2S-A = B+C$,$2S-B = A+C$,and $2S-C = A+B$. The expression becomes $\sin(B+C) + \sin(A+C) + \sin(A+B) - \sin(2S)$. Using $\sin(B+C) = \sin(2S-A)$,the expression is $\sin(2S-A) + \sin(2S-B) + \sin(2S-C) - \sin(2S)$. Using the identity $\sin X + \sin Y + \sin Z - \sin(X+Y+Z) = 4 \sin \frac{X+Y}{2} \sin \frac{Y+Z}{2} \sin \frac{Z+X}{2}$,Let $X = 2S-A$,$Y = 2S-B$,$Z = 2S-C$. Then $X+Y+Z = 6S - (A+B+C) = 6S - 2S = 4S$. This does not match the standard identity directly. However,for $A+B+C = 2S$,the identity $\sin A + \sin B + \sin C - \sin(A+B+C) = 4 \sin \frac{A+B-C}{2} \sin \frac{B+C-A}{2} \sin \frac{C+A-B}{2}$ is not applicable. Actually,for $A+B+C = 2S$,$\sin(2S-A) + \sin(2S-B) + \sin(2S-C) - \sin(2S) = 4 \sin \frac{A}{2} \sin \frac{B}{2} \sin \frac{C}{2}$.

If $A+B+C=2S$,then $\sin(2S-A)+\sin(2S-B)+\sin(2S-C)-\sin(2S) = $

Similar Questions

The ratio of the maximum and minimum values attained by the function $f(x) = 1 + 2 \sin x + 3 \cos^2 x$ for $0 \leq x \leq \frac{2\pi}{3}$ is

What is the maximum value of the function $f(x) = \sin x + \cos x$?

The range of $f(x) = \cos x - \sin x$ is

If $A+B=\frac{\pi}{2}$,then the maximum value of $\cos A \cdot \cos B$ is

If $u = \sqrt {{a^2}{{\cos }^2}\theta + {b^2}{{\sin }^2}\theta } + \sqrt {{a^2}{{\sin }^2}\theta + {b^2}{{\cos }^2}\theta } $,then the difference between the maximum and minimum values of ${u^2}$ is given by

Vedclass Test Series

Exam Paper Generator

Online Exam Module