Use the suggested substitution to write the expression as a trigonometric expression. Simplify your answer as much as possible. Assume #0 <= theta <= 2pi#. ?

Question

#sqrt (64-4x^2 #, #x/4=cos(theta)#

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Answer 1 · 2018-03-05T05:01:15

# 8sintheta#.

#x/4=costheta :. x=4costheta#.

#"Now, "sqrt(64-4x^2)=sqrt{4(16-x^2)}#,

#=sqrt4*sqrt(16-x^2)#,

#=2sqrt{16-(4costheta)^2}#,

#=2sqrt(16-16cos^2theta)#,

#=2sqrt{16(1-cos^2theta)}#,

#=2*sqrt16*sqrt{1-(costheta)^2}#,

#=2*4*sqrt{(sintheta)^2}#,

#=8|sintheta|#,

#=8(+sintheta),...[because, 0lethetalepi/2rArr sinthetage0]#,

#=8sintheta#.

Answer 2 · 2018-03-05T05:03:15

Given: #sqrt(64-4x^2)#

Remove 64 from under the radical and it becomes 8:

#sqrt(64-4x^2)= 8sqrt(1-x^2/16)#

#sqrt(64-4x^2)= 8sqrt(1-(x/4)^2)#

Substitute #x/4=cos(theta)#:

#sqrt(64-4x^2)= 8sqrt(1-(cos(theta))^2)#

#sqrt(64-4x^2)= 8sqrt(1-cos^2(theta))#

Substitute #sin(theta) = sqrt(1-cos^2(theta))#:

#sqrt(64-4x^2)= 8sin(theta)#