What is the sine for an angle that has a cosine of (2)/(5) and is in Quadrant IV?

Use the Pythagorean identity \sin^2 \theta + \cos^2 \theta = 1 and the quadrant to solve. Explain your reasoning.

1 answer

To find the sine of an angle \( \theta \) that has a cosine value of \( \cos \theta = \frac{2}{5} \) and lies in Quadrant IV, we can use the Pythagorean identity:

\[ \sin^2 \theta + \cos^2 \theta = 1. \]

First, we know \( \cos \theta = \frac{2}{5} \). We can substitute this value into the Pythagorean identity. Calculating \( \cos^2 \theta \):

\[ \cos^2 \theta = \left(\frac{2}{5}\right)^2 = \frac{4}{25}. \]

Substituting this into the Pythagorean identity gives:

\[ \sin^2 \theta + \frac{4}{25} = 1. \]

Next, we can isolate \( \sin^2 \theta \):

\[ \sin^2 \theta = 1 - \frac{4}{25}. \]

To perform the subtraction, we convert \( 1 \) into a fraction with the same denominator:

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

so we have:

\[ \sin^2 \theta = \frac{25}{25} - \frac{4}{25} = \frac{21}{25}. \]

Now, take the square root of both sides:

\[ \sin \theta = \pm \sqrt{\frac{21}{25}}. \]

This simplifies to:

\[ \sin \theta = \pm \frac{\sqrt{21}}{5}. \]

Since the angle \( \theta \) is in Quadrant IV and the sine function is negative in this quadrant, we take the negative root:

\[ \sin \theta = -\frac{\sqrt{21}}{5}. \]

Thus, the sine of the angle is:

\[ \boxed{-\frac{\sqrt{21}}{5}}. \]