Parabolas Worksheet .pdf - Pre-Calculus Day 1 Parabolas HW ... - Free Printable

Problem Analysis:

The task involves identifying and solving a problem related to the Pythagorean Identity in trigonometry. The Pythagorean Identity states:

\[
\sin^2(\theta) + \cos^2(\theta) = 1
\]

This identity is fundamental and can be used to solve various trigonometric problems, such as finding one trigonometric function given another and the quadrant of the angle.

From the image, we are tasked with solving for $\sin(\theta)$ given $\cos(\theta) = -\frac{4}{5}$ and the fact that $\theta$ lies in Quadrant III. Let us break this down step by step.

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Step 1: Recall the Pythagorean Identity

The Pythagorean Identity is:

\[
\sin^2(\theta) + \cos^2(\theta) = 1
\]

We are given:
\[
\cos(\theta) = -\frac{4}{5}
\]

Substitute $\cos(\theta)$ into the identity:

\[
\sin^2(\theta) + \left(-\frac{4}{5}\right)^2 = 1
\]

---

Step 2: Simplify the Equation

Calculate $\left(-\frac{4}{5}\right)^2$:

\[
\left(-\frac{4}{5}\right)^2 = \frac{16}{25}
\]

Substitute this back into the equation:

\[
\sin^2(\theta) + \frac{16}{25} = 1
\]

Isolate $\sin^2(\theta)$ by subtracting $\frac{16}{25}$ from both sides:

\[
\sin^2(\theta) = 1 - \frac{16}{25}
\]

Convert $1$ to a fraction with a denominator of $25$:

\[
1 = \frac{25}{25}
\]

So:

\[
\sin^2(\theta) = \frac{25}{25} - \frac{16}{25} = \frac{9}{25}
\]

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Step 3: Solve for $\sin(\theta)$

Take the square root of both sides:

\[
\sin(\theta) = \pm \sqrt{\frac{9}{25}}
\]

Simplify the square root:

\[
\sin(\theta) = \pm \frac{3}{5}
\]

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Step 4: Determine the Correct Sign for $\sin(\theta)$

Since $\theta$ is in Quadrant III, both sine and cosine are negative in this quadrant. Therefore, we choose the negative value for $\sin(\theta)$:

\[
\sin(\theta) = -\frac{3}{5}
\]

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Final Answer:

\[
\boxed{-\frac{3}{5}}
\]