91Ó°ÊÓ

From the triangle,

$\left|OA\right|=1$

For the triangle, the sum of the three interior angles is $180°$

Consider $∆OAC,$

$$\begin{gathered} ∠AOC+∠OCA+∠OAC=180° \\ \mathrm{Î\pm }+90°+∠OAC=180° \\ ∠OAC=90°-\mathrm{Î\pm } \end{gathered}$$

By using law of sines,

$\frac{\sin A}{a}=\frac{\sin B}{b}$

Substitute,

$$\begin{gathered} OC=b, \\ ∠OCA=90°, \\ \left|OA\right|=Afora, \\ ∠OAC=90°-\mathrm{Î\pm } \end{gathered}$$

$$\begin{gathered} \frac{\sin 90°}{1}=\frac{\sin (90°-\mathrm{Î\pm })}{\left|OC\right|} \\ \left|OC\right|=\cos \mathrm{Î\pm } \end{gathered}$$

Find the area of the triangle by using the formula,

$$\begin{gathered} K=\frac{1}{2}ab\sin C \\ =\frac{1}{2}\left(\left|OA\right|\right)\left(\left|OC\right|\right)\sin \mathrm{Î\pm } \\ =\frac{1}{2}\left(1\right)\cos \mathrm{Î\pm }.\sin \mathrm{Î\pm } \end{gathered}$$

Find the area by using the formula,

$$\begin{gathered} K=\frac{1}{2}ab.\sin C \\ =\frac{1}{2}\left|OC\right|\left|OB\right|\sin \mathrm{Î\pm } \\ =\frac{1}{2}\left|OB\right|\cos \mathrm{Î\pm }.\sin \mathrm{Î\pm } \end{gathered}$$

$$\begin{gathered} ∠COB+∠OBC+∠OCB=180° \\ \mathrm{β}+∠OBC+90°=180° \\ ∠OBC=90°-\mathrm{β} \end{gathered}$$

$$\begin{gathered} \frac{\sin (90°-\mathrm{β})}{\cos \mathrm{Î\pm }}=\frac{\sin 90°}{OB} \\ \left|OB\right|\cos \mathrm{β}=\cos \mathrm{Î\pm } \end{gathered}$$

Substitute the above equation to find the area,

$$\begin{gathered} K=\frac{1}{2}\left|OB\right|\left|OB\right|\cos \mathrm{β}.\sin \mathrm{β} \\ =\frac{1}{2}{\left|OB\right|}^2\cos \mathrm{β}.\sin \mathrm{β} \end{gathered}$$

From the figure,

$∠AOB=\mathrm{Î\pm }+\mathrm{β}$

Substitute

$$\begin{gathered} a=\left|OA\right| \\ =1 \\ b=\left|OB\right| \\ C=\mathrm{Î\pm }+\mathrm{β} \end{gathered}$$in the formula, we get,

$$\begin{gathered} K=\frac{1}{2}\left|OA\right|\left|OB\right|\sin (\mathrm{Î\pm }+\mathrm{β}) \\ =\frac{1}{2}\left(1\right)\left|OB\right|\sin (\mathrm{Î\pm }+\mathrm{β}) \end{gathered}$$

The law of sines is $\frac{\sin A}{a}=\frac{\sin B}{b}$

Substitute,

$$\begin{gathered} b=\left|OB\right| \\ a=\left|OC\right| \\ =\cos \mathrm{Î\pm } \\ A=∠OBC \\ =90°-\mathrm{β} \\ B=∠OCB \\ =90° \end{gathered}$$

$$\begin{gathered} \frac{\sin (90°-\mathrm{β})}{\cos \mathrm{Î\pm }}=\frac{\sin 90°}{\left|OB\right|} \\ \left|OB\right|\cos \mathrm{β}=\cos \mathrm{Î\pm } \end{gathered}$$

And hence

$\left|OB\right|=\frac{\cos \mathrm{Î\pm }}{\cos \mathrm{β}}$

From the figure,

Area of $∆OAB=$Area of $∆OAC+$Area of $∆OCB$

Substitute the known values,

$\frac{1}{2}\left|OB\right|\sin (\mathrm{Î\pm }+\mathrm{β})=\frac{1}{2}\sin \mathrm{Î\pm }.\cos \mathrm{Î\pm }+\frac{1}{2}{\left|OB\right|}^2\sin \mathrm{β}.\cos \mathrm{β}$

From the previous proofs,

$\left|OB\right|=\frac{\cos \mathrm{Î\pm }}{\cos \mathrm{β}},so,\cos \mathrm{Î\pm }=\left|OB\right|\cos \mathrm{β};\cos \mathrm{β}=\frac{\cos \mathrm{Î\pm }}{\left|OB\right|}$

$$\begin{gathered} \frac{1}{2}\left|OB\right|\sin (\mathrm{Î\pm }+\mathrm{β})=\frac{1}{2}\sin \mathrm{Î\pm }.\left|OB\right|\cos \mathrm{β}+\frac{1}{2}{\left|OB\right|}^2\sin \mathrm{β}.\frac{\cos \mathrm{Î\pm }}{\left|OB\right|} \\ =\frac{1}{2}\left|OB\right|\sin \mathrm{Î\pm }.\cos \mathrm{β}+\frac{1}{2}\left|OB\right|\sin \mathrm{β}.\cos \mathrm{Î\pm } \end{gathered}$$

Divide the whole equation by, $\frac{1}{2}\left|OB\right|$,

$\sin (\mathrm{Î\pm }+\mathrm{β})=\sin \mathrm{Î\pm }.\cos \mathrm{β}+\sin \mathrm{β}.\cos \mathrm{Î\pm }$