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Use the trigonometric substitution to write the algebraic expression as a trigonometric fuction of θ, where 0 < θ < Π/2

 

√(9 - x^2) , x = 3cos θ

 

answer: 3 sin θ

how???

 

thank you so much for helping!!

Guest Apr 23, 2017

Best Answer 

 #1
avatar+7155 
+3

\(\sqrt{9-x^2}\)

 

Replace x with 3cos θ .

\(=\sqrt{9-(3\cos \theta)^2} \\~\\ =\sqrt{9-9\cos^2 \theta}\)

 

Factor out a 9.

\(=\sqrt{9(1-\cos^2 \theta)}\)

 

The Pythagorean identity tells us that 1 - cos2θ = sin2θ

\(=\sqrt{9(\sin^2\theta)} \\~\\ =\sqrt9\cdot\sqrt{\sin^2\theta} =3\sin\theta\)

hectictar  Apr 23, 2017
 #1
avatar+7155 
+3
Best Answer

\(\sqrt{9-x^2}\)

 

Replace x with 3cos θ .

\(=\sqrt{9-(3\cos \theta)^2} \\~\\ =\sqrt{9-9\cos^2 \theta}\)

 

Factor out a 9.

\(=\sqrt{9(1-\cos^2 \theta)}\)

 

The Pythagorean identity tells us that 1 - cos2θ = sin2θ

\(=\sqrt{9(\sin^2\theta)} \\~\\ =\sqrt9\cdot\sqrt{\sin^2\theta} =3\sin\theta\)

hectictar  Apr 23, 2017

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