Why are those equations true?

Question

I am learning a heavy paper, and I attached the figure and the equations that the author claim.

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Answer 1

Hints:

• let $F'$ be the foot of the perpendicular from $F$ onto $MM_3$, then $MM_3=F'M_3-F'M$;

• use the law of sines in $\triangle MM_3F$.

Answer 2

Here is a more algebraic one. It looks big, but it is easy. There is just a lot of explanation.

For simplicity I will set $a=||{MM}_3||,\ b=||FM||$ and $c=||{FM}_3||$.

For the first one.

By the cosine rule in the triangle ${FMM}_3$ we get:

$$a^2=b^2+c^2-2bc\cos δ_1 (1)$$

We also have that:

$$\cos δ_1=\cos (\pi -(δ_2+δ_3))=-\cos (δ_2+δ_3)=-\cos δ_2\cos δ_3+\sin δ_2\sin δ_3$$

and so after substitution into $(1)$ we have:

$$a^2=b^2+c^2+2bc\cos δ_2\cos δ_3-2bc\sin δ_2\sin δ_3\Rightarrow \\ a^2=b^2(\cos^2 δ_3+\sin^2 δ_3)+c^2(\cos^2 δ_2+\sin^2 δ_2)+2bc\cos δ_3\cos δ_2-2bc\sin δ_3\sin δ_2\Rightarrow \\ a^2=(b\cos δ_3+c\cos δ_2)^2+(b\sin δ_3-c\sin δ_2)^2$$

The term $(b\sin δ_3-c\sin δ_2)^2$ vanishes, because the sine law in the above triangle implies that $\displaystyle{\frac{b}{\sin δ_2}=\frac{c}{\sin δ_3}}$ and this is equivalent to $b\sin δ_3=c\sin δ_2$.

Hence:

$$a^2=(b\cos δ_3+c\cos δ_2)^2\Rightarrow a=b\cos δ_3+c\cos δ_2\ \text{or}\ a=-b\cos δ_3-c\cos δ_2$$

We will reject the case $a=-b\cos δ_3-c\cos δ_2$. In contrary suppose that this is the case. Then if we apply the cosine rule twice, we have:

$$2a^2=-2ab\cos δ_3-2ac\cos δ_2\Rightarrow 2a^2=(c^2-a^2-b^2)+(b^2-a^2-c^2)\Rightarrow a=0,$$ contradiction. Therefore $a=b\cos δ_3+c\cos δ_2.$

For the second one, just apply the sine law again.