Problem 108 - 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 \(\cos \theta\). $$ -5 \sqrt{3}=\sqrt{100-x^{2}}, \quad x=10 \cos \theta $$

Step-by-Step Solution

Answer

\(\sin \theta = \frac{\sqrt{3}}{2}, \cos \theta = \frac{1}{2}\)

Step 1: Trigonometric Substitution

Substitute \(x\) with \(10 \cos \theta\) into the given algebraic equation. This gives us \(-5 \sqrt{3} = \sqrt{100 - (10 \cos \theta)^2}\).

Step 2: Simplification

Simplify the equation. We square both sides to get rid of the square root. Then simplify the equation to express \(\cos \theta\). This provides \(-5 \sqrt{3} = \sqrt{100 - (100 \cos^2 \theta)}\). Square both sides to get: \(75 = 100(1 - \cos^2 \theta)\). Arranging for \(\cos \theta\), we get: \(\cos \theta = \sqrt{1 - \frac{3}{4}} = \frac{1}{2}\).

Step 3: Find sin theta

We use the identity \(\sin^2 \theta + \cos^2 \theta = 1\), and simplify to get \(\sin \theta\). Substituting \(\cos \theta = \frac{1}{2}\), we get \(\sin \theta = \sqrt{1 - \left(\frac{1}{2}\right)^2} = \frac{\sqrt{3}}{2}\).

Step 4: Value Assignments

We check the quadrant \(\theta\) lies in based on the trigonometric substitutions and the original equation. \(x = 10 \cos \theta\) implies we are looking at values in the first and fourth quadrant. Given the values we found, \(\theta\) must be in the first quadrant, since both sine and cosine are positive there.