Let's solve this problem using the knowledge of trigonometric functions on the unit circle.
1. [tex]\(\sin(90^\circ)\)[/tex]
- On the unit circle, when [tex]\(\theta = 90^\circ\)[/tex], the point [tex]\(P\)[/tex] corresponds to [tex]\((\cos(90^\circ), \sin(90^\circ))\)[/tex].
- For [tex]\(\theta = 90^\circ\)[/tex], the point lies at the top of the unit circle, which is the point [tex]\((0, 1)\)[/tex].
- Therefore, [tex]\(\sin(90^\circ) = 1.0\)[/tex].
2. [tex]\(\cos(180^\circ)\)[/tex]
- When [tex]\(\theta = 180^\circ\)[/tex], the point [tex]\(P\)[/tex] corresponds to [tex]\((\cos(180^\circ), \sin(180^\circ))\)[/tex].
- For [tex]\(\theta = 180^\circ\)[/tex], the point lies at the leftmost point of the unit circle, which is the point [tex]\((-1, 0)\)[/tex].
- Therefore, [tex]\(\cos(180^\circ) = -1.0\)[/tex].
3. [tex]\(\sin(40^\circ)\)[/tex]
- [tex]\(\theta = 40^\circ\)[/tex] is not one of the standard angles, so we can't place this exactly using just special triangles and the unit circle.
- However, the numerical result can be obtained: [tex]\(\sin(40^\circ) \approx 0.6427876096865393\)[/tex].
4. [tex]\(\cos(245^\circ)\)[/tex]
- Similarly, [tex]\( \theta = 245^\circ\)[/tex] is also not a standard angle on the unit circle.
- But, from the numerical result, we know: [tex]\(\cos(245^\circ) \approx -0.42261826174069916\)[/tex].
Putting it all together:
[tex]\[
\begin{array}{c}
\sin \left(90^{\circ}\right)=1.0 \\
\cos \left(180^{\circ}\right)=-1.0 \\
\sin \left(40^{\circ}\right) \approx 0.6427876096865393 \\
\cos \left(245^{\circ}\right) \approx -0.42261826174069916 \\
\end{array}
\][/tex]
These are the values for the trigonometric functions based on the points on the unit circle.
