The Law of Cosines and the Law of Sines Are Equivalent

Sidney H. Kung
Cupertino, CA
December 2016

We show by using the projection property of triangle that the Law of Sines and the Law of Cosines are equivalent. Refer to the figure below:

Sine Law and Cosine Law are equivalent

$\alpha+\beta+\gamma=\pi,\,$ $a=b\cos\gamma+c\cos\beta.\,$ (Note that the latter holds even when one of the angles $\beta,\gamma\,$ is obtuse.) By squaring,

$\begin{align} a^2 &= b^2\cos^2\gamma+c^2\cos^2\beta+2bc\cos\beta\cos\gamma\\ &=b^2(1-\sin^2\gamma)+c^2(1-\sin^2\beta)+2bc\cos\beta\cos\gamma\\ &=b^2+c^2+2bc[\cos\beta\cos\gamma-\sin\beta\sin\gamma]\\ &\qquad\qquad-[b^2\sin^2\gamma+c^2\sin^2\beta-2bc\sin\beta\sin\gamma]\\ &=[b^2+c^2+2bc\cos (\beta+\gamma)]-(b\sin\gamma-c\sin\beta)^2\\ &=[b^2+c^2-2bc\cos \alpha]-(b\sin\gamma-c\sin\beta)^2, \end{align}$

where we used the Addition Formula for Cosine. Thus,


$a^2=[b^2+c^2-2bc\cos (\alpha)]-(b\sin\gamma-c\sin\beta)^2.$

This expression indicates that the Law of Sines and the Law of Cosines are equivalent in that, the two conditions $a^2=b^2+c^2-2bc\cos (\alpha)\,$ and $b\sin\gamma-c\sin\beta=0\,$ either hold or do not hold simultaneously, the latter being equivalent to the Law of Sines $\displaystyle \frac{\sin\beta}{b}=\frac{\sin\gamma}{c},\,$ the former is the expression of the Law of Cosines.

There have been articles giving, separately, a proof for the Laws of Sines and Cosines or a proof of the Law of Sines using the Law of Cosines ([1],[2],[3]). But with (1) we can do both simultaneously.


  1. Law of cosines, Sec 3.7 Using the law of sines,
  2. Patrik Nystedt, A proof of the law of sines using the law of cosines, to appear, 2017 Mathematics Magazine
  3. W. W. Sawyer, Prelude to Mathematics, Dover, 2011, pp 37-39


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