in triangle ABC, 4CosACosB is equal to
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Answer (1)
22nd Aug, 2020
In a triangle ABC cos(A+B)= cosAcosB-sinAsinB. cos(A-B)= cosAcosB+sinAsinB.
there by, by adding these two and multiplying the result by 2 times you get the desired answer. 4cosAcosB = 2[cos(A+B) + cos (A-B)]. This can also be done by interchanging the signs and the answer will remain the same.
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