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

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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

how???

thank you so much for helping!!

Guest Apr 23, 2017

#1
+7324
+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
+7324
+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