To solve for the other trigonometric values given [tex]\(\sin \theta\)[/tex] when [tex]\(\frac{\pi}{2} < \theta < \pi\)[/tex], we start with the Pythagorean identity [tex]\(\sin^2 \theta + \cos^2 \theta = 1\)[/tex].
1. We know that in the interval [tex]\(\frac{\pi}{2} < \theta < \pi\)[/tex], [tex]\(\sin \theta\)[/tex] is positive and [tex]\(\cos \theta\)[/tex] is negative.
2. The Pythagorean identity [tex]\(\sin^2 \theta + \cos^2 \theta = 1\)[/tex] equates to:
[tex]\[
\cos^2 \theta = 1 - \sin^2 \theta
\][/tex]
3. Taking the square root of both sides provides two possible solutions for [tex]\(\cos \theta\)[/tex]:
[tex]\[
\cos \theta = \pm\sqrt{1 - \sin^2 \theta}
\][/tex]
However, since [tex]\(\cos \theta\)[/tex] is negative in the interval [tex]\(\frac{\pi}{2} < \theta < \pi\)[/tex], we have:
[tex]\[
\cos \theta = -\sqrt{1 - \sin^2 \theta}
\][/tex]
4. Once we have found the values of [tex]\(\sin \theta\)[/tex] and [tex]\(\cos \theta\)[/tex], we can determine all other trigonometric ratios using their definitions and the known values.
To summarize, the correct explanation is:
- The values of [tex]\(\sin \theta\)[/tex] and [tex]\(\cos \theta\)[/tex] represent the legs of a right triangle with a hypotenuse of 1. Since [tex]\(\theta\)[/tex] is in Quadrant II, [tex]\(\cos \theta\)[/tex] is negative. Therefore, solving for [tex]\(\cos \theta\)[/tex] finds the unknown leg, and then all other trigonometric values can be found.
Therefore, the best option is:
- The values of [tex]\(\sin \theta\)[/tex] and [tex]\(\cos \theta\)[/tex] represent the legs of a right triangle with a hypotenuse of 1; therefore, solving for [tex]\(\cos \theta\)[/tex] finds the unknown leg, and then all other trigonometric values can be found.
