Question

If sinx cosy = 1/4 and 3tanx=4tany then 16sin(x-y)=?

Answer 1

Step-by-step explanation:

We have,

$\sin(x) \cos(y) = \frac{1}{4}$

And,

$3 \tan(x) = 4 \tan(y)$

$\implies \: \frac{\tan(x)}{ \tan(y) } = \frac{4}{3}$

$\implies \: \frac{\sin(x) \cos(y) }{ \cos(x) \sin(y) } = \frac{4}{3}$

$\implies \: \frac{ \frac{1}{4} }{ \cos(x) \sin(y) } = \frac{4}{3}$

$\implies \: \frac{1}{4} = \frac{4}{3} \cos(x) \sin(y)$

$\implies \: \cos(x) \sin(y) = \frac{3}{16}$

Now,

$16 \sin(x - y) = 16 \sin(x) \cos(y) - 16 \cos(x) \sin(y)$

$\implies16 \sin(x - y) = 16 \times \frac{1}{4} - 16 \times \frac{3}{16}$

$\implies16 \sin(x - y) = 4 - 1$

$\implies16 \sin(x - y) = 3$