# How do you use trigonometric substitution to write the algebraic expression sqrt(64-16x^2) as a trigonometric function of theta where 0<theta<pi/2 and x=2costheta?

Jan 2, 2017

$\sqrt{64 - 16 {x}^{2}} = 8 \sin \theta = 8 \sin \left(\arccos \left(\frac{x}{2}\right)\right)$

#### Explanation:

We know that $x = 2 \cos \theta$, and that ${x}^{2} = x \left(x\right)$.

=sqrt(64 - 16(2costheta)(2costheta)

=sqrt(64 - 16(4cos^2theta)

$= \sqrt{64 - 64 {\cos}^{2} \theta}$

$= \sqrt{64 \left(1 - {\cos}^{2} \theta\right)}$

We now use the identity ${\sin}^{2} x + {\cos}^{2} x = 1$ to solve the problem.

$= \sqrt{64 {\sin}^{2} \theta}$

$= \pm 8 \sin \theta$

However, since this is in the first quadrant, the expression must be positive.

$= 8 \sin \theta$

This can be rewritten with respect to $x$.

$x = 2 \cos \theta$

$\frac{x}{2} = \cos \theta$

$\theta = \arccos \left(\frac{x}{2}\right)$

Thus:

$= 8 \sin \left(\arccos \left(\frac{x}{2}\right)\right)$

Hopefully this helps!