# How do you simplify sin(2cos^-1(4/5))?

Nov 15, 2017

Use the identity $\sin \left(2 A\right) = 2 \sin \left(A\right) \cos \left(A\right)$
Then use the identity $\sin \left(A\right) = \pm \sqrt{1 - {\cos}^{2} \left(A\right)}$

#### Explanation:

Given: $\sin \left(2 {\cos}^{-} 1 \left(\frac{4}{5}\right)\right)$

Use the identity $\sin \left(2 A\right) = 2 \sin \left(A\right) \cos \left(A\right)$

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

The cosine of its inverse yields its argument:

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

Perform the multiplication:

$\frac{8}{5} \sin \left({\cos}^{-} 1 \left(\frac{4}{5}\right)\right)$

Use the identity #sin(A) = +-sqrt(1-cos^2(A))

$\pm \frac{8}{5} \sqrt{1 - {\cos}^{2} \left({\cos}^{-} 1 \left(\frac{4}{5}\right)\right)}$

Again, the cosine of its inverse yields its argument:

$\pm \frac{8}{5} \sqrt{1 - {\left(\frac{4}{5}\right)}^{2}}$

$\pm \frac{8}{5} \sqrt{\frac{25}{25} - \frac{16}{25}}$

$\sin \left(2 {\cos}^{-} 1 \left(\frac{4}{5}\right)\right) = \pm \frac{24}{25}$

To determine whether to choose the positive or negative value, one would need know whether the angle was in the first or fourth quadrant.