Answer :
Let the angle be defined by
[tex]$$
\theta = \cos^{-1}\left(\frac{14}{15}\right),
$$[/tex]
so that
[tex]$$
\cos \theta = \frac{14}{15}.
$$[/tex]
We want to find the exact values of
[tex]$$
\sin\left(\cos^{-1}\left(\frac{14}{15}\right)\right) = \sin \theta \quad \text{and} \quad \tan\left(\cos^{-1}\left(\frac{14}{15}\right)\right) = \tan \theta.
$$[/tex]
Follow these steps:
1. Use the Pythagorean Identity:
We know from the Pythagorean identity that
[tex]$$
\sin^2\theta + \cos^2\theta = 1.
$$[/tex]
Since [tex]$\cos \theta = \frac{14}{15}$[/tex], then
[tex]$$
\cos^2\theta = \left(\frac{14}{15}\right)^2 = \frac{196}{225}.
$$[/tex]
Subtract this from 1 to find [tex]$\sin^2 \theta$[/tex]:
[tex]$$
\sin^2\theta = 1 - \frac{196}{225} = \frac{225}{225} - \frac{196}{225} = \frac{29}{225}.
$$[/tex]
2. Determine [tex]$\sin \theta$[/tex]:
Since the range of [tex]$\cos^{-1}$[/tex] is [tex]$[0,\pi]$[/tex], and [tex]$\cos \theta = \frac{14}{15}$[/tex] is positive, the angle [tex]$\theta$[/tex] lies in the first quadrant where sine is also positive. Thus, we take the positive square root:
[tex]$$
\sin \theta = \sqrt{\frac{29}{225}} = \frac{\sqrt{29}}{15}.
$$[/tex]
3. Find [tex]$\tan \theta$[/tex]:
Recall that
[tex]$$
\tan\theta = \frac{\sin \theta}{\cos \theta}.
$$[/tex]
Substitute the values we found for [tex]$\sin \theta$[/tex] and [tex]$\cos \theta$[/tex]:
[tex]$$
\tan \theta = \frac{\frac{\sqrt{29}}{15}}{\frac{14}{15}} = \frac{\sqrt{29}}{14}.
$$[/tex]
4. Final Answers:
Therefore, the exact values are
[tex]$$
\sin\left(\cos^{-1}\left(\frac{14}{15}\right)\right) = \frac{\sqrt{29}}{15}
$$[/tex]
and
[tex]$$
\tan\left(\cos^{-1}\left(\frac{14}{15}\right)\right) = \frac{\sqrt{29}}{14}.
$$[/tex]
These are the exact values of the given trigonometric functions.
[tex]$$
\theta = \cos^{-1}\left(\frac{14}{15}\right),
$$[/tex]
so that
[tex]$$
\cos \theta = \frac{14}{15}.
$$[/tex]
We want to find the exact values of
[tex]$$
\sin\left(\cos^{-1}\left(\frac{14}{15}\right)\right) = \sin \theta \quad \text{and} \quad \tan\left(\cos^{-1}\left(\frac{14}{15}\right)\right) = \tan \theta.
$$[/tex]
Follow these steps:
1. Use the Pythagorean Identity:
We know from the Pythagorean identity that
[tex]$$
\sin^2\theta + \cos^2\theta = 1.
$$[/tex]
Since [tex]$\cos \theta = \frac{14}{15}$[/tex], then
[tex]$$
\cos^2\theta = \left(\frac{14}{15}\right)^2 = \frac{196}{225}.
$$[/tex]
Subtract this from 1 to find [tex]$\sin^2 \theta$[/tex]:
[tex]$$
\sin^2\theta = 1 - \frac{196}{225} = \frac{225}{225} - \frac{196}{225} = \frac{29}{225}.
$$[/tex]
2. Determine [tex]$\sin \theta$[/tex]:
Since the range of [tex]$\cos^{-1}$[/tex] is [tex]$[0,\pi]$[/tex], and [tex]$\cos \theta = \frac{14}{15}$[/tex] is positive, the angle [tex]$\theta$[/tex] lies in the first quadrant where sine is also positive. Thus, we take the positive square root:
[tex]$$
\sin \theta = \sqrt{\frac{29}{225}} = \frac{\sqrt{29}}{15}.
$$[/tex]
3. Find [tex]$\tan \theta$[/tex]:
Recall that
[tex]$$
\tan\theta = \frac{\sin \theta}{\cos \theta}.
$$[/tex]
Substitute the values we found for [tex]$\sin \theta$[/tex] and [tex]$\cos \theta$[/tex]:
[tex]$$
\tan \theta = \frac{\frac{\sqrt{29}}{15}}{\frac{14}{15}} = \frac{\sqrt{29}}{14}.
$$[/tex]
4. Final Answers:
Therefore, the exact values are
[tex]$$
\sin\left(\cos^{-1}\left(\frac{14}{15}\right)\right) = \frac{\sqrt{29}}{15}
$$[/tex]
and
[tex]$$
\tan\left(\cos^{-1}\left(\frac{14}{15}\right)\right) = \frac{\sqrt{29}}{14}.
$$[/tex]
These are the exact values of the given trigonometric functions.
