Problem 106 - Precalculus with Limits

Question

In Exercises \(105-108,\) use the trigonometric substitution to write the algebraic equation as a trigonometric equation of \(\theta\) where \(-\pi / 2<\theta<\pi / 2 .\) Then find \(\sin \theta\) and \(\cos \theta\). $$ 3=\sqrt{36-x^{2}}, \quad x=6 \sin \theta $$

Step-by-Step Solution

Answer

\(\sin\theta = \sqrt{3}/2\) and \(\cos\theta = 0.5\)

Step 1: Perform the substitution

Perform the substitution representing \(x = 6\sin\theta\) into \(3 = \sqrt{36 - x^2}\). So, the algebraic equation becomes \(3 = \sqrt{36 - 36\sin^2\theta}\).

Step 2: Simplify the equation

Simplify the equation. The expression under square root can be written as a difference of squares. Hence, \(3 = \sqrt{(6^2 - (6\sin\theta)^2)}\). This simplifies the equation to \(3 = 6\cos\theta\).

Step 3: Solve for θ

Solve for θ by dividing 6 on both sides of equation. This will yield \(\cos\theta = 0.5\). The angle \(\theta\) which yield \(\cos\theta = 0.5\) in the interval \(-\pi/2 < \theta < \pi/2\) is \(\theta = \pi/3\).

Step 4: Compute \(\sin\theta\) and \(\cos\theta\)

Using the obtained value of θ = \(\pi/3\) in \(\sin\theta\) and \(\cos\theta\). So, \(\sin\theta = \sin(\pi/3) = \sqrt{3}/2\) and \(\cos\theta = \cos(\pi/3) = 0.5\) as computed in the previous step.