Use the suggested substitution to write the expression as a trigonometric expression. Simplify your answer as much as possible. Assume 0 ≤ θ ≤ π2. ?

Question

#\sqrt(100-4x^(2)),(x)/(5)=sin(\theta )#

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Answer 1 · 2018-02-22T01:42:22

#sqrt(100-4x^2) = 10cos(theta)# where #x/5 = sin(theta)#.

#sqrt(100-4x^2)#

#x/5 = sin(theta)#

We may first rewrite the second equation in terms of #x# in order to substitute it into the first.

#color(red)x = color(red)(5sin(theta))#

#sqrt(100-4color(red)x^2)#

# = sqrt(100 - 4(5sin(theta))^2)#

# = sqrt(100 - 4(25sin^2(theta)))#

# = sqrt(100 - 100sin^2(theta))#

Conveniently, we are able to factor out 100 and simplify the radical.

# = sqrt(100*(1 - sin^2(theta)))#

# = sqrt(100)*sqrt(1 - sin^2(theta))#

# = 10color(blue)(sqrt(1 - sin^2(theta)))#

By the Pythagorean identity:

#sin^2(theta) + cos^2(theta) = 1#

#cos^2(theta) = 1 - sin^2(theta)#

#color(blue)(cos(theta)) = color(blue)(sqrt(1 - sin^2(theta)))#

Therefore,

#10color(blue)(sqrt(1 - sin^2(theta)))#

# = 10cos(theta)#