B.4 Cosine and Sine Laws
- Page ID
- 89666
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Cosine Law or Law of Cosines
The cosine law says that, if a triangle has sides of length \(a\text{,}\) \(b\) and \(c\) and the angle opposite the side of length \(c\) is \(\gamma\text{,}\) then
\begin{align*} c^2 &= a^2+b^2 - 2ab\cos\gamma \end{align*}
Observe that, when \(\gamma=\tfrac{\pi}{2}\text{,}\) this reduces to, (surpise!) Pythagoras' theorem \(c^2=a^2+b^2\text{.}\) Let's derive the cosine law.
Consider the triangle on the left. Now draw a perpendicular line from the side of length \(c\) to the opposite corner as shown. This demonstrates that
\begin{align*} c &= a \cos \beta + b \cos \alpha\\ \end{align*}
Multiply this by \(c\) to get an expression for \(c^2\text{:}\)
\begin{align*} c^2 &= ac \cos \beta + bc \cos \alpha\\ \end{align*}
Doing similarly for the other corners gives
\begin{align*} a^2 &= ac \cos \beta + ab \cos \gamma\\ b^2 &= bc \cos \alpha + ab \cos \gamma \end{align*}
Now combining these:
\begin{align*} a^2+b^2-c^2 &= (bc-bc) \cos \alpha + (ac-ac)\cos\beta + 2ab \cos \gamma\\ &= 2ab\cos \gamma \end{align*}
as required.
Sine Law or Law of Sines
The sine law says that, if a triangle has sides of length \(a, b\) and \(c\) and the angles opposite those sides are \(\alpha\text{,}\) \(\beta\) and \(\gamma\text{,}\) then
\begin{align*} \frac{a}{\sin \alpha} &= \frac{b}{\sin \beta} = \frac{c}{\sin \gamma}. \end{align*}
This rule is best understood by computing the area of the triangle using the formula \(A = \frac{1}{2}ab\sin\theta\) of Appendix A.10. Doing this three ways gives
\begin{align*} 2A &= bc \sin \alpha\\ 2A &= ac \sin \beta\\ 2A &= ab \sin \gamma \end{align*}
Dividing these expressions by \(abc\) gives
\begin{align*} \frac{2A}{abc} &= \frac{\sin \alpha}{a} = \frac{\sin\beta}{b} = \frac{\sin \gamma}{c} \end{align*}
as required.