Question

Using the Pythagorean Identity, determine cosθ
if sinθ=−1217
and π<θ<3π2
.(1 point)
Responses

−145√17
negative Start Fraction Start Root 145 End Root over 17 End Fraction

−2917−−√
negative Start Root Start Fraction 29 over 17 End Fraction End Root

2917−−√
Start Root Start Fraction 29 over 17 End Fraction End Root

145√17

Answer 1

To determine \(\cos \theta\) given \(\sin \theta = -\frac{12}{17}\) and \(\pi < \theta < \frac{3\pi}{2}\), we can use the Pythagorean identity:

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

First, compute \(\sin^2 \theta\):

\[
\sin^2 \theta = \left(-\frac{12}{17}\right)^2 = \frac{144}{289}.
\]

Now, substitute \(\sin^2 \theta\) into the Pythagorean identity:

\[
\frac{144}{289} + \cos^2 \theta = 1.
\]

To isolate \(\cos^2 \theta\):

\[
\cos^2 \theta = 1 - \frac{144}{289}.
\]

To combine the fractions, convert 1 to have a common denominator:

\[
1 = \frac{289}{289},
\]

therefore,

\[
\cos^2 \theta = \frac{289}{289} - \frac{144}{289} = \frac{145}{289}.
\]

Taking the square root of both sides gives us:

\[
\cos \theta = \pm \sqrt{\frac{145}{289}} = \pm \frac{\sqrt{145}}{17}.
\]

Now, we need to determine the sign of \(\cos \theta\). We are given that \(\pi < \theta < \frac{3\pi}{2}\), which is in the third quadrant. In the third quadrant, the cosine function is negative. Therefore, we conclude that:

\[
\cos \theta = -\frac{\sqrt{145}}{17}.
\]

The correct response is:

\[
-\frac{\sqrt{145}}{17}.
\]