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# Consider the statement given below. State if this statement is true or false.${{a}^{2}}\cos 2B+{{b}^{2}}\cos 2A+2ab\cos (A-B)={{c}^{2}}$A. TrueB. False  Verified
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Hint: Take the LHS of the expression given. By using the formulas of $\cos 2\theta$and$\cos (A-B)$, simplify the expression. Use the projection formula of the triangle to solve the rest. In the projection formula, the length of any side of a triangle is equal to the sum of the projections of the other 2 sides.

We have been given the expression,
${{a}^{2}}\cos 2B+{{b}^{2}}\cos 2A+2ab\cos (A-B)={{c}^{2}}.....(1)$
We know the basic trigonometric formula,
$\cos 2\theta ={{\cos }^{2}}\theta -{{\sin }^{2}}\theta$
It can be written as
$\cos 2\theta ={{\cos }^{2}}\theta -(1-{{\cos }^{2}}\theta )={{\cos }^{2}}\theta -1+{{\cos }^{2}}\theta =2{{\cos }^{2}}\theta -1$.
We know that \begin{align} & {{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1 \\ & \Rightarrow {{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta \\ \end{align}
The expression of $\cos (A-B)=\cos A\cos B+\sin A\sin B.$
Now let us substitute the values of $\cos 2\theta$and$\cos (A-B)$ in LHS of equation (1).
\begin{align} & \cos 2\theta ={{\cos }^{2}}\theta -{{\sin }^{2}}\theta \\ & \therefore \cos 2B={{\cos }^{2}}B-{{\sin }^{2}}B \\ & \cos 2A={{\cos }^{2}}A-{{\sin }^{2}}A \\ \end{align}
\begin{align} & LHS={{a}^{2}}\cos 2B+{{b}^{2}}\cos 2A+2ab\cos (A-B) \\ & ={{a}^{2}}({{\cos }^{2}}B-si{{n}^{2}}B)+{{b}^{2}}({{\cos }^{2}}A-{{\sin }^{2}}A)+2ab\left[ \cos A\cos B+\sin A\sin B \right] \\ \end{align}
Let us open the brackets and simplify it.
\begin{align} & ={{a}^{2}}{{\cos }^{2}}B-{{a}^{2}}{{\sin }^{2}}B+{{b}^{2}}{{\cos }^{2}}A-{{b}^{2}}{{\sin }^{2}}A+2ab.\cos A.\cos B+2ab.\sin A.\sin B \\ & =({{a}^{2}}{{\cos }^{2}}B+{{b}^{2}}{{\cos }^{2}}A+2ab.\cos A.cosB)-({{a}^{2}}{{\sin }^{2}}B+{{b}^{2}}{{\sin }^{2}}A-2ab.\sin A.\sin B) \\ \end{align}
They are of the form ${{a}^{2}}+2ab+{{b}^{2}}={{(a+b)}^{2}}.$
${{(a\cos B+b\cos A)}^{2}}-{{(a\sin B-b\sin A)}^{2}}$
By projection formulae, the length of any side of a triangle is equal to the sum of the projections of the other two sides.
$\therefore a\cos B+b\cos A=C$for any triangle ABC.
\begin{align} & {{(a\cos B+b\cos A)}^{2}}-{{(a\sin nB+b\sin A)}^{2}} \\ & ={{c}^{2}}-{{(a\sin B-b\sin A)}^{2}} \\ \end{align}
By sine formula, we know that $\dfrac{\sin A}{a}=\dfrac{sinB}{b}$
Cross multiplying the above expression, we get,
\begin{align} & b\sin A=a\sin B \\ & ={{c}^{2}}-{{(a\sin B-b\sin A)}^{2}} \\ & ={{c}^{2}}-{{(a\sin B-a\sin B)}^{2}} \\ & ={{c}^{2}}-0={{c}^{2}} \\ \end{align}
Therefore, we proved that,
${{a}^{2}}\cos 2B+{{b}^{2}}\cos 2A+2ab\cos (A-B)={{c}^{2}}$.
Thus the statement is true.
Option A is the correct answer.

Note: Here we use the cosine formulae to solve the expression. You should remember the formulae so that solving questions like these would become easy. We used the projection formula which is an important concept, to solve the expression.
$c=a\cos B+b\cos A$.