Question
given cos theta=-5/6 and angle theta is in quadrant 2 what is the exact value of sin theta in simplest form simplify all radicals if needed
Answers
GPT-4o mini
Given that \(\cos \theta = -\frac{5}{6}\) and \(\theta\) is in the second quadrant, we can use the Pythagorean identity to find \(\sin \theta\). The Pythagorean identity states:
\[
\sin^2 \theta + \cos^2 \theta = 1
\]
Substituting the given value of \(\cos \theta\):
\[
\sin^2 \theta + \left(-\frac{5}{6}\right)^2 = 1
\]
Calculating \(\left(-\frac{5}{6}\right)^2\):
\[
\sin^2 \theta + \frac{25}{36} = 1
\]
Next, we isolate \(\sin^2 \theta\):
\[
\sin^2 \theta = 1 - \frac{25}{36}
\]
To perform the subtraction, we need a common denominator. The number \(1\) can be written as \(\frac{36}{36}\):
\[
\sin^2 \theta = \frac{36}{36} - \frac{25}{36} = \frac{11}{36}
\]
Now, we take the square root of both sides:
\[
\sin \theta = \pm \sqrt{\frac{11}{36}} = \pm \frac{\sqrt{11}}{6}
\]
Since \(\theta\) is in the second quadrant, where sine is positive, we have:
\[
\sin \theta = \frac{\sqrt{11}}{6}
\]
Thus, the exact value of \(\sin \theta\) in simplest form is:
\[
\boxed{\frac{\sqrt{11}}{6}}
\]
\[
\sin^2 \theta + \cos^2 \theta = 1
\]
Substituting the given value of \(\cos \theta\):
\[
\sin^2 \theta + \left(-\frac{5}{6}\right)^2 = 1
\]
Calculating \(\left(-\frac{5}{6}\right)^2\):
\[
\sin^2 \theta + \frac{25}{36} = 1
\]
Next, we isolate \(\sin^2 \theta\):
\[
\sin^2 \theta = 1 - \frac{25}{36}
\]
To perform the subtraction, we need a common denominator. The number \(1\) can be written as \(\frac{36}{36}\):
\[
\sin^2 \theta = \frac{36}{36} - \frac{25}{36} = \frac{11}{36}
\]
Now, we take the square root of both sides:
\[
\sin \theta = \pm \sqrt{\frac{11}{36}} = \pm \frac{\sqrt{11}}{6}
\]
Since \(\theta\) is in the second quadrant, where sine is positive, we have:
\[
\sin \theta = \frac{\sqrt{11}}{6}
\]
Thus, the exact value of \(\sin \theta\) in simplest form is:
\[
\boxed{\frac{\sqrt{11}}{6}}
\]