91Ó°ÊÓ

We are given a figure.

$$\begin{gathered} f\left(x\right)=\sin x \\ g\left(x\right)=\cos x,\text{and} \\ h\left(x\right)=\tan x \end{gathered}$$

It can be seen that $(x,1)$is a point on the graph of the function, $x^2+y^2=4$. Substituting 1 for y and solve the equation, we get

$$\begin{gathered} x^2+1^2=4 \\ {x}^2+1=4 \\ x=\sqrt{3} \end{gathered}$$

So, we get the side opposite to the angle $\mathrm{Î\pm }$as $1$, the adjacent side as $\sqrt{3}$, and the hypotenuse as $2$.

So, $\sin \mathrm{Î\pm }=\frac{1}{2}$and

$\cos \mathrm{Î\pm }=\frac{\sqrt{3}}{2}$

Similarly, for second figure $\left(\frac{1}{3},y\right)$is a point on the graph of $x^2+y^2=1$. Substituting $x=\frac{1}{3}$and solve the equation, we get

localid="1646587549462" $$\begin{gathered} {\left(\frac{1}{3}\right)}^2+y^2=1 \\ \frac{1}{9}+y^2=1 \\ {y}^2=1-\frac{1}{9} \\ y=\sqrt{\frac{8}{9}} \\ y=Â\pm \frac{{\displaystyle 2\sqrt{2}}}{3} \\ y=-\frac{{\displaystyle 2\sqrt{2}}}{3} \end{gathered}$$

Because it is in the fourth quadrant.

From the figure, we have $\mathrm{θ}=2\mathrm{π}-\mathrm{β}$. So,

localid="1646626905980" $$\begin{gathered} \sin \mathrm{θ}=\sin (2\mathrm{π}-\mathrm{β}) \\ \sin (2\mathrm{π}-\mathrm{β})=\sin (-\mathrm{β}) \\ =-\sin \mathrm{β} \end{gathered}$$

and

localid="1646626911894" $$\begin{gathered} \cos \mathrm{θ}=\cos (2\mathrm{π}-\mathrm{β}) \\ \cos (2\mathrm{π}-\mathrm{β})=\cos (-\mathrm{β}) \\ =\cos \mathrm{β} \end{gathered}$$

So,

localid="1646627320456" role="math" $$\begin{gathered} \sin \mathrm{β}=-\frac{2\sqrt{2}}{3} \\ \cos \mathrm{β}=\frac{1}{3} \end{gathered}$$

The value of function is given by,

$$\begin{gathered} f(\mathrm{Î\pm }+\mathrm{β})=\sin (\mathrm{Î\pm }+\mathrm{β}) \\ =\sin \mathrm{Î\pm }\cos \mathrm{β}+\cos \mathrm{Î\pm }\sin \mathrm{β} \\ =\frac{1}{2}·\frac{1}{3}+\frac{\sqrt{3}}{2}\left(-\frac{2\sqrt{2}}{3}\right) \\ =\frac{1}{6}+\left(-\frac{2\sqrt{6}}{6}\right) \\ =\left(\frac{1-2\sqrt{6}}{6}\right) \end{gathered}$$