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Question: If \(A,B,C\) are angles of a triangle, then \(\sin 2A + \sin 2B - \sin 2C\) is equal to...

If A,B,CA,B,C are angles of a triangle, then sin⁡2A+sin⁡2B−sin⁡2C\sin 2A + \sin 2B - \sin 2C is equal to

A

4sin⁡Acos⁡Bcos⁡C4\sin A\cos B\cos C

B

4cos⁡A4\cos A

C

4sin⁡Acos⁡A4\sin A\cos A

D

4cos⁡Acos⁡Bsin⁡C4\cos A\cos B\sin C

Answer

4cos⁡Acos⁡Bsin⁡C4\cos A\cos B\sin C

Explanation

Solution

sin⁡2A+sin⁡2B−sin⁡2C=2sin⁡Acos⁡A+2cos⁡(B+C)sin⁡(B−C)[∵A+B+C=π,B+C=π−A,cos⁡(B+C)=\sin 2A + \sin 2B - \sin 2C = 2\sin A\cos A + 2\cos(B + C)\sin(B - C)\lbrack\because A + B + C = \pi,B + C = \pi - A,\cos(B + C) = $$\cos(\pi - A),\cos(B + C) = - \cos A,\sin(B + C) = \sin A\rbrack

= 2cos⁡A[sin⁡A−sin⁡(B−C)]2\cos A\lbrack\sin A - \sin(B - C)\rbrack = 2cos⁡A[sin⁡(B+C)−sin⁡(B−C)]2\cos A\lbrack\sin(B + C) - \sin(B - C)\rbrack

= 2cos⁡A.2cos⁡B.sin⁡C=4cos⁡A.cos⁡B.sin⁡C2\cos A.2\cos B.\sin C = 4\cos A.\cos B.\sin C

Trick: First put A=B=C=60oA = B = C = 60^{o}, for these values. Options (1) and (2) satisfies the condition.

Now put A=B=45oA = B = 45^{o} and C=90oC = 90^{o}, then only (4) satisfies.

Hence (4) is the answer.