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Question: If in a triangle ABC, ∠C =90<sup>0</sup>, then the maximum value of sinAsinB is...

If in a triangle ABC, ∠C =900, then the maximum value of sinAsinB is

A

12\frac{1}{2}

B

1

C

2

D

None of these

Answer

12\frac{1}{2}

Explanation

Solution

sinA sinB = 12×2sin⁡Asin⁡B\frac{1}{2} \times 2\sin A\sin B

12[cos⁡(A−B)−cos⁡(A+B)]\frac{1}{2}\left\lbrack \cos(A - B) - \cos(A + B) \right\rbrack

=12[cos⁡(A−B)−cos⁡90∘]\frac{1}{2}\left\lbrack \cos(A - B) - \cos 90{^\circ} \right\rbrack=12cos⁡(A−B)\frac{1}{2}\cos(A - B) ≤ 12\frac{1}{2}

⇒ Maximum value of sinA sinB =12\frac{1}{2}