Use trigonometric substitution to write the algebraic expression as a trigonometric function of θ, where 0 < θ?
Use trigonometric substitution to write the algebraic expression as a trigonometric function of θ, where 0 < θ < π/2.
√(100-4x²) , x = 5cosθ
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Verified answer
Just plug-n-chug.
√(100 - 4x²) = √(100 - 4(25cos²Θ)) = √(100 - 100cos²Θ)
=10√(1 - cos²Θ) = 10√(sin²Θ) = 10|sinΘ|.
Since 0 < Θ < π/2, the sine is positive. So this simplifies to 10sinΘ.
If 0 < θ < π/2, then sin θ is positive. So
√(100-4x²) =
√(100-4(5cos(θ))²) =
√(100 - 100 cos²(θ)) =
10 √(1 - cos²(θ)) =
10 √sin²(θ) =
10 sin θ
sqrt[8 - 2x^2] , x= 2cos? x= 2cos? x^2 = 4cos^2(x) 2x^2 = 8cos^2(x) sqrt(8 - 2x^2) sqrt[8 - 8cos^2(x)] sqrt[8(a million - cos^2(x))] aspect word: sin^2(x) + cos^2(x) = a million so sin^2(x) = a million - cos^2(x) making use of the "aspect word": sqrt[8(a million - cos^2(x))] sqrt[8sin^2(x)] sqrt(8)sin(x) 2sqrt(2)sin(x) <==== answer