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