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

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

Uniteasy Admin · Accepted Answer

$$\sin A\sin B = \cos^2 \dfrac{C}{2}$$
$$\sin A\sin B = \dfrac{\cos C+1}{2} $$
$$\because C = \pi-(A+B) $$
$$	herefore  2\sin A\sin B = -\cos(A+B)+1$$
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

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