MathsGee is Zero-Rated (You do not need data to access) on: Telkom |Dimension Data | Rain | MWEB

0 like 0 dislike
12 views
Use complex numbers to prove the Law of Cosine: Let $\triangle A B C$ be a triangle with
\begin{aligned} |B C|=& a,|C A|=b,|A B|=c \text { and } \angle B C A=\theta . \text { Then } \\ & a^{2}+b^{2}-2 a b \cos \theta=c^{2} \end{aligned}
Hint: Place $C$ at the origin, $B$ at $z_{1}$ and $A$ at $z_{2}$. Prove that
$$z_{1} \bar{z}_{2}+z_{2} \bar{z}_{1}=2\left|z_{1} z_{2}\right| \cos \theta$$
| 12 views

0 like 0 dislike
Proof. Following the hint, we let $C=0, B=z_{1}$ and $A=z_{2} .$ Then $a=\left|z_{1}\right|, b=\left|z_{2}\right|$
\text { and } \begin{aligned} c=\left|z_{2}-z_{1}\right| . \text { So } \\ \qquad \begin{aligned} a^{2}+b^{2}-c^{2} &=\left|z_{1}\right|^{2}+\left|z_{2}\right|^{2}-\left|z_{2}-z_{1}\right|^{2} \\ &=\left(z_{1} \bar{z}_{1}+z_{2} \bar{z}_{2}\right)-\left(z_{2}-z_{1}\right) \overline{\left(z_{2}-z_{1}\right)} \\ &=\left(z_{1} \bar{z}_{1}+z_{2} \bar{z}_{2}\right)-\left(z_{2}-z_{1}\right)\left(\bar{z}_{2}-\bar{z}_{1}\right) \\ &=\left(z_{1} \bar{z}_{1}+z_{2} \bar{z}_{2}\right)-\left(z_{1} \bar{z}_{1}+z_{2} \bar{z}_{2}-z_{1} \bar{z}_{2}-z_{2} \bar{z}_{1}\right) \\ &=z_{1} \bar{z}_{2}+z_{2} \bar{z}_{1} \end{aligned} \end{aligned}
Let $z_{1}=r_{1} e^{i \theta_{1}}$ and $z_{2}=r_{2} e^{i \theta_{2}} .$ Then
\begin{aligned} z_{1} \bar{z}_{2}+z_{2} \bar{z}_{1} &=r_{1} e^{i \theta_{1}} \overline{r_{2} e^{i \theta_{2}}}+r_{2} e^{i \theta_{2}} \overline{r_{1} e^{i \theta_{1}}} \\ &=\left(r_{1} e^{i \theta_{1}}\right)\left(r_{2} e^{-i \theta_{2}}\right)+\left(r_{2} e^{i \theta_{2}}\right)\left(r_{1} e^{-i \theta_{1}}\right) \\ &=r_{1} r_{2} e^{i\left(\theta_{1}-\theta_{2}\right)}+r_{1} r_{2} e^{i\left(\theta_{2}-\theta_{1}\right)} \\ &=2 r_{1} r_{2} \cos \left(\theta_{1}-\theta_{2}\right)=2\left|z_{1}\right|\left|z_{2}\right| \cos \theta=2 a b \cos \theta \end{aligned}

Therefore, we have
$$a^{2}+b^{2}-c^{2}=z_{1} \bar{z}_{2}+z_{2} \bar{z}_{1}=2 a b \cos \theta$$
and hence
$$a^{2}+b^{2}-2 a b \cos \theta=c^{2}$$
by Gold Status (10,261 points)

0 like 0 dislike