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Question

Express the Law of Sines in terms of only the cosine function.

Solution

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According to law of sines:

asina=bsinb=csinc\dfrac{a}{\sin a}=\dfrac{b}{\sin b}=\dfrac{c}{\sin c}

We know sin(90°+θ)=cosθ\sin (90 \text{\textdegree}+\theta)=\cos \theta.

Expressing law of sines in terms of cosines:

asina=bsinb=csincasin(90°90°+a)=bsin(90°90°+b)=csin(90°90°+c)acos(a90°)=bcos(b90°)=ccos(c90°)\begin{align*} \dfrac{a}{\sin a}&=\dfrac{b}{\sin b}=\dfrac{c}{\sin c}\\ \dfrac{a}{\sin (90\text{\textdegree} -90\text{\textdegree} + a)}&=\dfrac{b}{\sin(90\text{\textdegree} -90\text{\textdegree} + b)}=\dfrac{c}{\sin(90\text{\textdegree} -90\text{\textdegree} + c)}\\ \textcolor{#4257b2}{\dfrac{a}{\cos (a-90\text{\textdegree} )}}&=\textcolor{#4257b2}{\dfrac{b}{\cos( b-90\text{\textdegree} )}=\dfrac{c}{\cos( c-90\text{\textdegree} )}}\\ \end{align*}

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