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

Question

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#?

Emery Adkins · Accepted Answer

#sqrt(64 - 16x^2) = 8sintheta = 8sin(arccos(x/2))#

We know that #x = 2costheta#, and that #x^2 = x(x)#.

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

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

#=sqrt(64 - 64cos^2theta)#

#=sqrt(64(1 - cos^2theta))#

We now use the identity #sin^2x + cos^2x = 1# to solve the problem.

#=sqrt(64sin^2theta)#

#= +-8sintheta#

However, the expression needs to be positive because this is in the first quadrant.

#= 8sintheta#

This can be rewritten with respect to #x#.

#x = 2costheta#

#x/2 = costheta#

#theta = arccos(x/2)#

Thus:

#= 8sin(arccos(x/2))#

I hope this is useful!

Brooklyn Anderson · Answer

undefined To express $ \sqrt{64 - 16x^2} $ as a trigonometric function of $ \theta $, where $ 0 < \theta < \frac{\pi}{2} $ and $ x = 2 \cos(\theta) $, follow these steps:

1. Rewrite $ x = 2 \cos(\theta) $ in terms of $ \theta $ to express $ x $ as a trigonometric function.
2. Substitute the expression for $ x $ into $ \sqrt{64 - 16x^2} $.
3. Simplify the expression using trigonometric identities.
4. Determine the appropriate trigonometric function of $ \theta $ that represents $ \sqrt{64 - 16x^2} $ in the given interval.

By following these steps, you can express $ \sqrt{64 - 16x^2} $ as a trigonometric function of $ \theta $ within the specified domain.

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

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#?