Question

Calculating the principal value, find the value of $\sin \left[2{\sin }^{-1}\left(\frac{4}{5}\right)\right]$.

Answer 1

We are given, $sin\left[2xi{n}^{-1}\left(\frac{4}{5}\right)\right]$

Let $si{n}^{-1}\left(\frac{4}{5}\right)=\theta$

where $\theta ϵ[-\frac{\pi }{2},\frac{\pi }{2}]$

So, $sin\theta =\frac{4}{5}$

now, $cos\theta =\sqrt{co{s}^2\theta }$

$=\sqrt{1-si{n}^2\theta }$

$cos\theta =\sqrt{1-{\left(\frac{4}{5}\right)}^2}$

$cos\theta =\sqrt{1-\frac{16}{25}}$

$cos\theta =\sqrt{\frac{9}{25}}=\pm \frac{3}{5}$

but $\theta ϵ[-\pi /2,\pi /2]$ so, $cos\theta =\frac{3}{5}$

$\therefore sin\left[2si{n}^{-1}\left(\frac{4}{5}\right)\right]$

$=sin2\theta$

$=2sin\theta cos\theta$

$=2\left(\frac{4}{5}\right)\left(\frac{3}{5}\right)$

$=\frac{24}{25}$

So, $sin\left[2si{n}^{-1}\frac{4}{5}\right]=\frac{24}{25}$