Trigonometry! #73

Geometry Level 2

If cos ⁡ ( A − B ) = 3 5 \cos(A-B)=\frac{3}{5} tan ⁡ A tan ⁡ B = 2 \tan A \tan B = 2 then the value of sin ⁡ A sin ⁡ B \sin A \sin B is

This problem is part of the set Trigonometry .


The answer is 0.4.

Omkar Kulkarni by Omkar Kulkarni , 22, India

This section requires Javascript.
You are seeing this because something didn't load right. We suggest you, (a) try refreshing the page, (b) enabling javascript if it is disabled on your browser and, finally, (c) loading the non-javascript version of this page . We're sorry about the hassle.

1 solution

Omkar Kulkarni
Feb 21, 2015

tan ⁡ A tan ⁡ B = 2 sin ⁡ A sin ⁡ B cos ⁡ A cos ⁡ B = 2 cos ⁡ A cos ⁡ B = sin ⁡ A sin ⁡ B 2 \tan A\tan B=2 \\ \frac{\sin A\sin B}{\cos A\cos B}=2 \\ \cos A\cos B=\frac{\sin A\sin B}{2}

Now,

cos ⁡ ( A − B ) = 3 5 cos ⁡ A cos ⁡ B + sin ⁡ A sin ⁡ B = 3 5 sin ⁡ A sin ⁡ B 2 + sin ⁡ A sin ⁡ B = 3 5 3 sin ⁡ A sin ⁡ B 2 = 3 5 sin ⁡ A sin ⁡ B = 2 5 \cos(A-B)=\frac{3}{5} \\ \cos A\cos B+\sin A\sin B=\frac{3}{5} \\ \frac{\sin A\sin B}{2}+\sin A\sin B=\frac{3}{5} \\ \frac{3\sin A\sin B}{2}=\frac{3}{5} \\ \boxed{\sin A\sin B=\frac{2}{5}}

2 Helpful 0 Interesting 0 Brilliant 0 Confused

0 pending reports

×

Problem Loading...

Note Loading...

Set Loading...