Outline simple geometric derivations of the
formulas for $\sin (\alpha+\beta)$ and $\cos (\alpha+\beta)$ in the case in which $\alpha$ and $\beta$ are acute angles, with $\alpha+\beta < 90^{\circ} .$ The exercises rely on the accompanying figures, which are constructed as follows. Begin, in Figure $A,$ with $\alpha=\angle G A D, \beta=\angle H A G,$ and $A H=1$ Then, from $H,$ draw perpendiculars to $\overline{A D}$ and to $\overline{A G},$ as shown in Figure $B .$ Finally, draw $\overline{F E} \perp \overline{B H}$ and $\overline{F C} \perp \overline{A D}$
(FIGURES CAN'T COPY)
Formula for $\sin (\alpha+\beta) .$ Supply the reasons or steps behind each statement.
(a) $B H=\sin (\alpha+\beta)$
(b) $F H=\sin \beta$
(c) $\angle B H F=\alpha$
(d) $E H=\cos \alpha \sin \beta \quad$ Hint: Use $\triangle E F H$ and the result in part (b).
(e) $A F=\cos \beta$
(f) $C F=\sin \alpha \cos \beta$
(g) $\sin (\alpha+\beta)=\sin \alpha \cos \beta+\cos \alpha \sin \beta$
Hint: $\sin (\alpha+\beta)=B H=E H+C F$