Question

Prove that
\[{{\cos }^{2}}A+{{\cos }^{2}}B+{{\cos }^{2}}C=1-2\cos A\cdot \cos B\cdot \cos C\]

Answer 1

Hint: We solve this question by first considering the left-hand side of the given expression and multiply and divide it with 2. Then we use the formula \[\cos (2\theta )=2{{\cos }^{2}}\theta -1\] and write the angles inside the trigonometric identities as multiples of 2. Then we use the formula \[\cos x+\cos y=2\cos \left( \dfrac{x+y}{2} \right)\cos \left( \dfrac{x-y}{2} \right)\] for simplifying the obtained expression above and then use the fact that \[A+B+C={{180}^{\circ }}\] and \[\cos ({{180}^{\circ }}-x)=-\cos x\] to simplify it further. Then we take $\cos C$ common from the expression and apply the formula \[\cos x-\cos y=2\sin \left( \dfrac{x+y}{2} \right)\sin \left( \dfrac{y-x}{2} \right)\] and then use the formula \[A+B+C={{180}^{\circ }}\] again to simplify the expression further and prove the given expression.

Complete step-by-step answer:
As mentioned in the question, we have to prove the given expression \[{{\cos }^{2}}A+{{\cos }^{2}}B+{{\cos }^{2}}C=1-2\cos A\cdot \cos B\cdot \cos C\].

Now, we will start with the left-hand side that is L.H.S. and try to make the necessary changes to convert it into the RHS.
\[\Rightarrow {{\cos }^{2}}A+{{\cos }^{2}}B+{{\cos }^{2}}C\]
Now multiplying and dividing the expression with 2, we get the following
\[\begin{align}
  & \Rightarrow {{\cos }^{2}}A+{{\cos }^{2}}B+{{\cos }^{2}}C \\
 & \Rightarrow \dfrac{2{{\cos }^{2}}A+2{{\cos }^{2}}B+2{{\cos }^{2}}C}{2} \\
\end{align}\]
Now let us consider the formula \[\cos 2x=2{{\cos }^{2}}x-1\]. Using it we can write the above equation as,
\[\begin{align}
  & \Rightarrow \dfrac{\cos 2A+1+\cos 2B+1+\cos 2C+1}{2} \\
 & \Rightarrow \dfrac{3+\cos 2A+\cos 2B+\cos 2C}{2} \\
\end{align}\]
Now using the formula \[\cos x+\cos y=2\cos \left( \dfrac{x+y}{2} \right)\cos \left( \dfrac{x-y}{2} \right)\], we can write as follows,
\[\begin{align}
  & \Rightarrow \dfrac{3+\cos 2A+\cos 2B+\cos 2C}{2} \\
 & \Rightarrow \dfrac{3+2\cos \left( \dfrac{2A+2B}{2} \right)\cos \left( \dfrac{2A-2B}{2} \right)+\cos 2C}{2} \\
 & \Rightarrow \dfrac{2+2\cos \left( A+B \right)\cos \left( A-B \right)+\cos 2C+1}{2} \\
 & \Rightarrow \dfrac{2+2\cos \left( A+B \right)\cos \left( A-B \right)+2{{\cos }^{2}}C}{2} \\
\end{align}\]
Now let consider the fact that,
\[A+B+C={{180}^{\circ }}\]
Hence, we can write as follows
\[\begin{align}
  & \Rightarrow \dfrac{2+2\cos \left( A+B \right)\cos \left( A-B \right)+2{{\cos }^{2}}C}{2} \\
 & \Rightarrow \dfrac{2+2\cos \left( {{180}^{\circ }}-C \right)\cos \left( A-B \right)+2{{\cos }^{2}}C}{2} \\
\end{align}\]
Now let us consider the formula, \[\cos \left( {{180}^{\circ }}-x \right)=-\cos x\]. Using it we can convert the above equation as,
\[\begin{align}
  & \Rightarrow \dfrac{2-2\cos C\cos \left( A-B \right)+2{{\cos }^{2}}C}{2} \\
 & \Rightarrow 1-\cos C\cos \left( A-B \right)+{{\cos }^{2}}C \\
 & \Rightarrow 1-\cos C\left( \cos \left( A-B \right)-\cos C \right) \\
\end{align}\]
Now using the formula \[\cos x-\cos y=2\sin \left( \dfrac{x+y}{2} \right)\sin \left( \dfrac{y-x}{2} \right)\], we can write as follows
\[\begin{align}
  & \Rightarrow 1-\cos C\left( 2\sin \left( \dfrac{A-B+C}{2} \right)\sin \left( \dfrac{C-A+B}{2} \right) \right) \\
 & \Rightarrow 1-2\cos C\left( \sin \left( \dfrac{A+C-B}{2} \right)\sin \left( \dfrac{\left( B+C \right)-A}{2} \right) \right) \\
\end{align}\]
Now let us consider \[A+B+C={{180}^{\circ }}\]. We can write it as,
\[A+C={{180}^{\circ }}-B\ and\ B+C={{180}^{\circ }}-A\]
Using these we can convert the above equation as,
\[\begin{align}
  & \Rightarrow 1-2\cos C\left( \sin \left( \dfrac{{{180}^{\circ }}-B-B}{2} \right)\sin \left( \dfrac{\left( {{180}^{\circ }}-A \right)-A}{2} \right) \right) \\
 & \Rightarrow 1-2\cos C\left( \sin \left( \dfrac{{{180}^{\circ }}-2B}{2} \right)\sin \left( \dfrac{{{180}^{\circ }}-2A}{2} \right) \right) \\
 & \Rightarrow 1-2\cos C\left( \sin \left( {{90}^{\circ }}-B \right)\sin \left( {{90}^{\circ }}-A \right) \right) \\
\end{align}\]
Now let us consider the formula,
\[\sin \left( {{90}^{\circ }}-A \right)=\cos A\]
Using this formula, we can write the above equation as,
\[\Rightarrow 1-2\cos C\cos B\cos A\]
So, we get that \[{{\cos }^{2}}A+{{\cos }^{2}}B+{{\cos }^{2}}C=1-2\cos A\cdot \cos B\cdot \cos C\]
Hence Proved.

Note: Another method of attempting this question is by converting the right-hand side, that is R.H.S. to the left hand side that is L.H.S. by using the relations that are given in the hint. Through this method also, we could get to the correct answer and hence, we would be able to prove the required proof. It is similar to the proof we have done but starts from the bottom of our proof and goes to the top.