Question
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.
Use the Pythagorean identity \sin^2 \theta + \cos^2 \theta = 1 and the quadrant to solve. Explain your reasoning.
Answers
GPT-4o mini
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:
\[
\boxed{-\frac{\sqrt{21}}{5}}.
\]
\[
\sin^2 \theta + \cos^2 \theta = 1.
\]
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:
\[
\boxed{-\frac{\sqrt{21}}{5}}.
\]