Solveeit Logo
Login

Question

Question: In triangle ABC, the value of \(\sin 2A + \sin 2B + \sin 2C\) is equal to...

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

A

4sin⁡A.sin⁡B.sin⁡C4\sin A.\sin B.\sin C

B

4cos⁡A.cos⁡B.cos⁡C4\cos A.\cos B.\cos C

C

2cos⁡A.cos⁡B.cos⁡C2\cos A.\cos B.\cos C

D

2sin⁡A.sin⁡B.sin⁡C2\sin A.\sin B.\sin C

Answer

4sin⁡A.sin⁡B.sin⁡C4\sin A.\sin B.\sin C

Explanation

Solution

We know that A+B+C=180∘A + B + C = 180{^\circ} (in ΔABC\Delta ABC)

Now, sin⁡2A+sin⁡2B+sin⁡2C\sin 2A + \sin 2B + \sin 2C

=2sin⁡(A+B)cos⁡(A−B)+2sin⁡Ccos⁡C= 2\sin(A + B)\cos(A - B) + 2\sin C\cos C

=2sin⁡(π−C)cos⁡(A−B)+2sin⁡Ccos⁡(π−A+B‾)= 2\sin(\pi - C)\cos(A - B) + 2\sin C\cos(\pi - \overline{A + B})

=2sin⁡Ccos⁡(A−B)−2sin⁡Ccos⁡(A+B)= 2\sin C\cos(A - B) - 2\sin C\cos(A + B)

=2sin⁡C{cos⁡(A−B)−cos⁡(A+B)}= 2\sin C\{\cos(A - B) - \cos(A + B)\}

=2sin⁡C{2sin⁡Asin⁡B}=4sin⁡Asin⁡Bsin⁡C= 2\sin C\{ 2\sin A\sin B\} = 4\sin A\sin B\sin C.