Question

Prove that $a\left( b\cos C- c\cos B \right)={{b}^{2}}-{{c}^{2}}$ where $a,b$ and $c$ are sides opposite to the angles with measure $A,B,C$ in triangle ABC.\[\]

Answer 1

Hint: We proceed from the left hand side of the statement of proof by using the projection formula$a=b\cos C+c\cos B$. We then use algebraic identity${{x}^{2}}-{{y}^{2}}=\left( x+y \right)\left( x-y \right)$, the trigonometric identity ${{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta $ and the sine law $\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}$ to arrive at the right hand side of the equation.

Complete step-by-step solution:
We are given the question that $a,b$, and $c$ are the length of sides opposite to the angles with measure $A, B, C$ in triangle ABC. We have the right figure of the problem below, \[\]
 

We observe in the above triangle that
\[\angle BAC=\angle A,\angle ABC=\angle B,\angle ACB=\angle C\]
We know from projection formula that
\[a=b\cos C+c\cos B\]
We know from sine law or rule that those lengths of sides of triangle are always in proportion with sines of opposite angles which we can write in symbol as
\[\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}\]
We take second and third term in the above equation to have,
\[\begin{align}
  & \dfrac{b}{\sin B}=\dfrac{c}{\sin C} \\
 & \Rightarrow b\sin C=c\sin B \\
 & \Rightarrow {{\left( b\sin C \right)}^{2}}={{\left( c\sin B \right)}^{2}} \\
 & \Rightarrow {{b}^{2}}{{\sin }^{2}}C={{c}^{2}}{{\sin }^{2}}B.....\left( 1 \right) \\
\end{align}\]
We are asked to prove the statement
\[ a\left( b\cos C- c\cos B \right)={{b}^{2}}-{{c}^{2}}\]
Let us proceed from the left hand side of the statement and use the projection formula $a=b\cos C+c\cos B$ to have,
\[ a\left( b\cos C- c\cos B \right)=\left( b\cos C+c\cos B \right)\left( b\cos C- c\cos B \right)\]
We use the algebraic identity ${{x}^{2}}-{{y}^{2}}=\left( x+y \right)\left( x-y \right)$ for $x=b\cos C,y=c\cos B$in the above step to have,
\[={{b}^{2}}{{\cos }^{2}}C-{{c}^{2}}{{\cos }^{2}}B\]
We use the trigonometric identity ${{\cos }^{2}}\theta =1-{{\sin }^{2}}\theta $ for any angle with measure $\theta $ for $\theta =C,B$in above step to have,
\[ \begin{align}
  & ={{b}^{2}}\left( 1-{{\sin }^{2}}C \right)-{{c}^{2}}\left( 1-{{\sin }^{2}}B \right) \\
 & ={{b}^{2}}-{{c}^{2}}-\left( {{b}^{2}}{{\sin }^{2}}C-{{c}^{2}}{{\sin }^{2}}B \right) \\
\end{align}\]
We use the value from (1) in the above step and proceed to have,
\[ \begin{align}
  & ={{b}^{2}}-{{c}^{2}}-\left( 0 \right) \\
 & ={{b}^{2}}-{{c}^{2}} \\
\end{align}\]
The above expression is at the right hand side of the statement of proof and hence the statement is proved.

Note: We can alternatively solve the above problem using law of cosine relates the sides of a triangle with cosine of one of the angles from where we obtain $\cos C=\dfrac{{{a}^{2}}+{{b}^{2}}-{{c}^{2}}}{2ab}$ and $\cos B=\dfrac{{{b}^{2}}+{{c}^{2}}-{{a}^{2}}}{2bc}$. We put $\cos B,\cos C$ on the left-hand side of the statement of proof, and simplify to arrive at the right-hand side. The law of sine and cosine is used to obtain the third side when lengths of two sides and angles are given in a technique known as triangulation.