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#?

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Answer 1 · 2017-01-02T16:12:38

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

#= 8sintheta#

This can be rewritten with respect to #x#.

#x = 2costheta#

#x/2 = costheta#

#theta = arccos(x/2)#

Thus:

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

Hopefully this helps!