In triangle ABC,the value of sin 2A + sin 2B | vedclass

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

Question

(A) In $\Delta ABC$,we have $A + B + C = 180^\circ$.Consider the expression $\sin 2A + \sin 2B + \sin 2C$.Using the sum-to-product formula $\sin x + \

Vedclass · Accepted Answer

$4\sin A \sin B \sin C$

Vedclass · Answer

(A) In $\Delta ABC$,we have $A + B + C = 180^\circ$.Consider the expression $\sin 2A + \sin 2B + \sin 2C$.Using the sum-to-product formula $\sin x + \sin y = 2\sin(\frac{x+y}{2})\cos(\frac{x-y}{2})$:$\sin 2A + \sin 2B = 2\sin(A+B)\cos(A-B)$.Since $A+B = 180^\circ - C$,then $\sin(A+B) = \sin C$.So,$\sin 2A + \sin 2B = 2\sin C \cos(A-B)$.Now,$\sin 2C = 2\sin C \cos C = 2\sin C \cos(180^\circ - (A+B)) = -2\sin C \cos(A+B)$.Substituting these back:$\sin 2A + \sin 2B + \sin 2C = 2\sin C \cos(A-B) - 2\sin C \cos(A+B)$.$= 2\sin C [\cos(A-B) - \cos(A+B)]$.Using the identity $\cos(A-B) - \cos(A+B) = 2\sin A \sin B$:$= 2\sin C [2\sin A \sin B] = 4\sin A \sin B \sin C$.

In triangle $ABC$,the value of $\sin 2A + \sin 2B + \sin 2C$ is equal to

Similar Questions

In $\triangle ABC$ with the usual notations,if $\left(\tan \frac{A}{2}\right)\left(\tan \frac{B}{2}\right)=\frac{3}{4}$,then $a+b=\ldots$ (in $c$)

In a $\Delta ABC$,$a = 5, b = 4$ and $\cos(A - B) = \frac{31}{32}$,then side $c$ is equal to

If in a $\triangle ABC$,$\frac{1}{a+c} + \frac{1}{b+c} = \frac{3}{a+b+c}$,then $\angle C$ is equal to (in $^{\circ}$)

Let $a, b$ and $c$ denote the lengths of sides $BC, CA$ and $AB$ of $\triangle ABC$. In $\triangle ABC$,$\angle BAC = 30^{\circ}$ and $\angle ABC = 60^{\circ}$. Then $a: b: c$ is

The area of a triangle is $10\sqrt{3} \text{ cm}^2$,angle $C = 60^{\circ}$,and its perimeter is $20 \text{ cm}$. Find the length of side $c$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module