# Thread: inverse trigonometry

1. ## inverse trigonometry

Show that 2sin-1(3/5) = sin-1(24/25)

any help is greatly appreciated ! thankyou !

2. Originally Posted by iiharthero
Show that 2sin-1(3/5) = sin-1(24/25)

any help is greatly appreciated ! thankyou !
it is the first time I solved something like this

$\displaystyle \sin ^{-1} \left(\frac{3}{5}\right) = u$

$\displaystyle \frac{3}{5} = \sin u$

$\displaystyle \cos u = \frac{4}{5}$

$\displaystyle \sin u \cos u = \frac{3}{5}\left(\frac{4}{5}\right)$

$\displaystyle \sin u \cos u = \frac{12}{25}$

$\displaystyle \frac{\sin 2u }{2} = \frac{12}{25}$

$\displaystyle \sin 2u = \frac{24}{25}$

$\displaystyle 2u=\sin ^{-1} \left(\frac{24}{25}\right)$

but
$\displaystyle u=\sin ^{-1} \left(\frac{3}{5}\right)$

$\displaystyle 2\sin ^{-1} \left(\frac{3}{5}\right)=\sin ^{-1} \left(\frac{24}{25}\right)$

3. Let $\displaystyle \frac{3}{5}=sin\theta$

$\displaystyle cos\theta=\frac{4}{5}$

$\displaystyle sin2\theta=2sin\theta cos\theta$

$\displaystyle sin2\theta=2*\frac{3}{5}*\frac{4}{5}$

$\displaystyle sin2\theta=\frac{24}{25}$

$\displaystyle 2\theta=sin^{-1}{\frac{24}{25}}$

$\displaystyle 2sin^{-1}{\frac{3}{5}}=sin^{-1}{\frac{24}{25}}$