Refer to the figure. If $\left|OA\right|=1$, show that:

(a) Area $∆OAC=\frac{1}{2}\sin \alpha .\cos \alpha$

(b) Area $∆OCB=\frac{1}{2}{\left|OB\right|}^2\sin \beta .\cos \beta$

(c) Area $∆OAB=\frac{1}{2}\left|OB\right|\sin (\alpha +\beta )$

(d) $\left|OB\right|=\frac{\cos \alpha }{\cos \beta }$

(e) $\sin (\alpha +\beta )=\sin \alpha .\cos \beta +\cos \alpha .\sin \beta$

From the triangle,

$\left|OA\right|=1$

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

Consider $∆OAC,$

$$\begin{gathered} \angle AOC+\angle OCA+\angle OAC=180° \\ \alpha +90°+\angle OAC=180° \\ \angle OAC=90°-\alpha \end{gathered}$$

By using law of sines,

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

Substitute,

$$\begin{gathered} OC=b, \\ \angle OCA=90°, \\ \left|OA\right|=Afora, \\ \angle OAC=90°-\alpha \end{gathered}$$

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

Find the area of the triangle by using the formula,

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

Find the area by using the formula,

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

$$\begin{gathered} \angle COB+\angle OBC+\angle OCB=180° \\ \beta +\angle OBC+90°=180° \\ \angle OBC=90°-\beta \end{gathered}$$

$$\begin{gathered} \frac{\sin (90°-\beta )}{\cos \alpha }=\frac{\sin 90°}{OB} \\ \left|OB\right|\cos \beta =\cos \alpha \end{gathered}$$

Substitute the above equation to find the area,

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

From the figure,

$\angle AOB=\alpha +\beta$

Substitute

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

$$\begin{gathered} K=\frac{1}{2}\left|OA\right|\left|OB\right|\sin (\alpha +\beta ) \\ =\frac{1}{2}\left(1\right)\left|OB\right|\sin (\alpha +\beta ) \end{gathered}$$

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

Substitute,

$$\begin{gathered} b=\left|OB\right| \\ a=\left|OC\right| \\ =\cos \alpha \\ A=\angle OBC \\ =90°-\beta \\ B=\angle OCB \\ =90° \end{gathered}$$

$$\begin{gathered} \frac{\sin (90°-\beta )}{\cos \alpha }=\frac{\sin 90°}{\left|OB\right|} \\ \left|OB\right|\cos \beta =\cos \alpha \end{gathered}$$

And hence

$\left|OB\right|=\frac{\cos \alpha }{\cos \beta }$

From the figure,

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

Substitute the known values,

$\frac{1}{2}\left|OB\right|\sin (\alpha +\beta )=\frac{1}{2}\sin \alpha .\cos \alpha +\frac{1}{2}{\left|OB\right|}^2\sin \beta .\cos \beta$

From the previous proofs,

$\left|OB\right|=\frac{\cos \alpha }{\cos \beta },so,\cos \alpha =\left|OB\right|\cos \beta ;\cos \beta =\frac{\cos \alpha }{\left|OB\right|}$

$$\begin{gathered} \frac{1}{2}\left|OB\right|\sin (\alpha +\beta )=\frac{1}{2}\sin \alpha .\left|OB\right|\cos \beta +\frac{1}{2}{\left|OB\right|}^2\sin \beta .\frac{\cos \alpha }{\left|OB\right|} \\ =\frac{1}{2}\left|OB\right|\sin \alpha .\cos \beta +\frac{1}{2}\left|OB\right|\sin \beta .\cos \alpha \end{gathered}$$

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

$\sin (\alpha +\beta )=\sin \alpha .\cos \beta +\sin \beta .\cos \alpha$