Question

What is the sine for an angle that has a cosine of 2/5 and is in Quadrant IV? Use the Pythagorean identity sin2θ+cos2θ=1 and the quadrant to solve.(1 point)

Answer 1

To find the sine of an angle given that its cosine is \( \frac{2}{5} \) and it is in Quadrant IV, we can use the Pythagorean identity:

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

Given that \( \cos \theta = \frac{2}{5} \), we can substitute this value into the identity.

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

Calculating \( \left( \frac{2}{5} \right)^2 \):

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

Now plug this back into the equation:

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

Now, subtract \( \frac{4}{25} \) from both sides:

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

To perform the subtraction, we need a common denominator. The number 1 can be expressed as \( \frac{25}{25} \):

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

Next, we take the square root of both sides to solve for \( \sin \theta \):

\[
\sin \theta = \pm \sqrt{\frac{21}{25}} = \pm \frac{\sqrt{21}}{5}
\]

Since the angle \( \theta \) is in Quadrant IV, where the sine value is negative, we have:

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

Thus, the sine for the angle is:

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