# How do you find the value of cos [2 Sin^-1 (-24/25)]?

May 17, 2015

For any angle $\theta$, we have $\cos 2 \theta = {\cos}^{2} \theta - {\sin}^{2} \theta$. We also know that ${\sin}^{2} \theta + {\cos}^{2} \theta = 1$.

So $\cos 2 \theta = {\cos}^{\theta} - {\sin}^{2} \theta$

$= \left(1 - {\sin}^{2} \theta\right) - {\sin}^{2} \theta$

$= 1 - {\sin}^{2} \theta$

If $\theta = {\sin}^{-} 1 \left(- \frac{24}{25}\right)$

then $\sin \theta = - \frac{24}{25}$

and $\cos 2 \theta = 1 - {\sin}^{2} \theta$

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

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

$= 1 - {24}^{2} / {25}^{2}$

$= \frac{{25}^{2} - {24}^{2}}{25} ^ 2$

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

$= \frac{\left(2 \times 24\right) + 1}{25} ^ 2$

$= \frac{49}{625}$