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Question: If \(\sin A = \sin B\)and \(\cos A = \cos B,\) then...

If sin⁡A=sin⁡B\sin A = \sin Band cos⁡A=cos⁡B,\cos A = \cos B, then

A

sin⁡A−B2=0\sin\frac{A - B}{2} = 0

B

sin⁡A+B2=0\sin\frac{A + B}{2} = 0

C

cos⁡A−B2=0\cos\frac{A - B}{2} = 0

D

cos⁡(A+B)=0\cos(A + B) = 0

Answer

sin⁡A−B2=0\sin\frac{A - B}{2} = 0

Explanation

Solution

We have sin⁡A=sin⁡B\sin A = \sin B and cos⁡A=cos⁡B\cos A = \cos B

sin⁡Asin⁡B=cos⁡Acos⁡B⇒sin⁡Acos⁡B−cos⁡Asin⁡B=0⇒sin⁡(A−B)=0\frac{\sin A}{\sin B} = \frac{\cos A}{\cos B} \Rightarrow \sin A\cos B - \cos A\sin B = 0 \Rightarrow \sin(A - B) = 0

Hence, sin⁡(A−B2)=0.\sin\left( \frac{A - B}{2} \right) = 0.