# How do you prove 10sin(x)cos(x)=6cos(x)?

May 19, 2018

If we simplify the equation by dividing both sides by $\cos \left(x\right)$, we obtain:

$10 \sin \left(x\right) = 6$, which implies
$\sin \left(x\right) = \frac{3}{5.}$

The right triangle which $\sin \left(x\right) = \frac{3}{5}$ is a 3:4:5 triangle, with legs $a = 3$, $b = 4$ and hypotenuse $c = 5$. From this we know that if $\sin \left(x\right) = \frac{3}{5}$ (opposite over hypotenuse), then $\cos = \frac{4}{5}$ (adjacent over hypotenuse). If we plug these identities back into the equation we reveal its validity:

$10 \left(\frac{3}{5}\right) \cdot \left(\frac{4}{5}\right) = 6 \left(\frac{4}{5}\right)$.

This simplifies to

$\frac{24}{5} = \frac{24}{5}$.

Therefore the equation is true for $\sin \left(x\right) = \frac{3}{5.}$