Question

In \triangle ABC, if \sin A\sin B = \cos^2 \dfrac{C}{2}, evaluate the type of the triangle

Collected in the board: Trigonometry

Steven Zheng posted 1 month ago


Answer

\sin A\sin B = \cos^2 \dfrac{C}{2}


\sin A\sin B = \dfrac{\cos C+1}{2}

\because C = \pi-(A+B)

\therefore 2\sin A\sin B = -\cos(A+B)+1

Using the sum and difference identity for cosine function,

2\sin A\sin B+\cos A\cos B -\sin A \sin B = 1

\cos A\cos B +\sin A \sin B = 1

\cos(A-B) = 1

Therefore, A= B, the triangle is an isoseles triangle

Steven Zheng posted 1 month ago

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