Question

If $A+B+C=\pi$, prove that ${{\cos }^{2}}A+{{\cos }^{2}}B-{{\cos }^{2}}C=1-2\sin A\sin B\cos C$.

Answer 1

Hint: For solving this question as there are squares of trigonometric ratios so, we will trigonometric formulas like formula for $\cos 2\theta$ and then further for simplifying the term written on the left-hand side we will use formulas for $\cos C+\cos D$, $\cos C-\cos D$ and $\cos \left( \pi -\theta \right)$. After that, we will prove the term on the left-hand side is equal to the term on the right-hand side.

Complete step-by-step answer:

Given:

It is given that if $A+B+C=\pi$ and we have to prove the following equation:

${{\cos }^{2}}A+{{\cos }^{2}}B-{{\cos }^{2}}C=1-2\sin A\sin B\cos C$

Now, before we proceed we should know the following four formulas:

$\begin{aligned}

& \cos 2\theta =2{{\cos }^{2}}\theta -1 \\

& \Rightarrow 2{{\cos }^{2}}\theta =\cos 2\theta +1 \\

& \Rightarrow {{\cos }^{2}}\theta =\dfrac{\cos 2\theta +1}{2}................................\left( 1 \right) \\

& \cos C+\cos D=2\cos \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right)..................\left( 2 \right) \\

& \cos C-\cos D=-2\sin \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{C-D}{2} \right).................\left( 3 \right) \\

& A+B+C=\pi \\

& \Rightarrow A+B=\pi -C \\

& \Rightarrow \cos \left( A+B \right)=\cos \left( \pi -C \right)=-\cos C.............................\left( 4 \right) \\

\end{aligned}$

Now, we will be using the above four formulas to simplify the term on the left-hand side to prove the desired result.

Now, L.H.S is equal to ${{\cos }^{2}}A+{{\cos }^{2}}B-{{\cos }^{2}}C$ so, using the formula from equation (1). Then,

$\begin{aligned}

& {{\cos }^{2}}A+{{\cos }^{2}}B-{{\cos }^{2}}C \\

& \Rightarrow \dfrac{1+\cos 2A}{2}+\dfrac{1+\cos 2B}{2}-{{\cos }^{2}}C \\

& \Rightarrow \dfrac{1}{2}+\dfrac{1}{2}+\dfrac{\left( \cos 2A+\cos 2B \right)}{2}-{{\cos }^{2}}C \\

& \Rightarrow 1+\dfrac{\left( \cos 2A+\cos 2B \right)}{2}-{{\cos }^{2}}C \\

\end{aligned}$

Now, using the formula from equation (2) in the above equation. Then,

$\begin{aligned}

& 1+\dfrac{\left( \cos 2A+\cos 2B \right)}{2}-{{\cos }^{2}}C \\

& \Rightarrow 1+\dfrac{1}{2}\left( 2\cos \left( \dfrac{2A+2B}{2} \right)\cos \left( \dfrac{2A-2B}{2} \right) \right)-{{\cos }^{2}}C \\

& \Rightarrow 1+\cos \left( A+B \right)\cos \left( A-B \right)-{{\cos }^{2}}C \\

\end{aligned}$

Now, using the formula from equation (4) in the above equation. Then,

$\begin{aligned}

& 1+\cos \left( A+B \right)\cos \left( A-B \right)-{{\cos }^{2}}C \\

& \Rightarrow 1-\cos c\cos \left( A-B \right)-{{\cos }^{2}}C \\

& \Rightarrow 1+\cos C\left( -\cos C-\cos \left( A-B \right) \right) \\

& \Rightarrow 1+\cos C\left( \cos \left( A+B \right)-\cos \left( A-B \right) \right) \\

\end{aligned}$

Now, using the formula from equation (3) in the above equation. Then,

$\begin{aligned}

& 1+\cos C\left( \cos \left( A+B \right)-\cos \left( A-B \right) \right) \\

& \Rightarrow 1+\cos C\left( -2\sin \left( \dfrac{A+B+A-B}{2} \right)\sin \left(

\dfrac{A+B-A+B}{2} \right) \right) \\

& \Rightarrow 1+\cos C\left( -2\sin A\sin B \right) \\

& \Rightarrow 1-2\sin A\sin B\cos C \\

\end{aligned}$

Now, from the above result we can say that ${{\cos }^{2}}A+{{\cos }^{2}}B-{{\cos }^{2}}C=1-2\sin A\sin B\cos C$.

Thus, $L.H.S=R.H.S$.

Hence Proved.

Note: Here, the student should first understand what we have to prove in the question and then proceed in a stepwise manner while solving. For making the simplification process smooth, we should also try to make use of trigonometric results like $\cos \left( \pi -\theta \right)=-\cos \theta$ for making equations that will help us further in the solution. Moreover, the formulas like $\cos C+\cos D$ , $\cos C-\cos D$ and $\cos 2\theta$ should be applied correctly with proper values and avoid making calculation mistakes while solving the problem.