Question

Prove that: \[{{\cos }^{2}}A+{{\cos }^{2}}B-2\cos A\cos B\cos \left( A+B \right)={{\sin }^{2}}\left( A+B \right)\]

Answer 1

Hint: For solving this question, first we apply cos(A+B) identity in the left-hand side. And apply all the trigonometry suitable formulas, we easily prove left hand side equal to right hand side.

Complete step-by-step answer:

Some of the useful trigonometric formula used in solving this problem:

cos (A + B) = cos A cos B − sin A sin B

sin (A + B) = sin A cos B + cos A sin B

${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1$

According to the problem statement, we consider the left-hand side of the equation for proving equivalence of both sides. First, we expand the left-hand side using the above-mentioned formula.

Considering the left-hand side of the question, we have

\[\Rightarrow {{\cos }^{2}}A+{{\cos }^{2}}B-2\cos A\cos B\cos \left( A+B \right)\]

Expand cos (A + B) by using the above identity, we get

\[\begin{align}

  & \Rightarrow {{\cos }^{2}}A+{{\cos }^{2}}B-2\cos A\cos B\left[ \cos A\cos B-\sin A\sin B \right] \\

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

\right)+2\left( \cos A \right)\left( \cos B \right)\left( \sin A \right)\left( \sin B \right) \\

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

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

\right)\left( \sin A \right)\left( \sin B \right) \\

\end{align}\]

Taking ${{\cos }^{2}}A$ common from 1 and 2 terms and ${{\cos }^{2}}B$ from 3 and 4 terms, we get

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

\right)+2\left( \cos A \right)\left( \cos B \right)\left( \sin A \right)\left( \sin B \right)$

We know the identity, ${{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1$. Therefore, $1-{{\cos }^{2}}\theta ={{\sin }^{2}}\theta $, using this we get

\[\begin{align}

  & \Rightarrow {{\cos }^{2}}A{{\sin }^{2}}B+{{\cos }^{2}}B{{\sin }^{2}}A+2\left[ \left( \cos A \right)\left( \sin B \right) \right]\left[ \left( \sin A \right)\left( \cos B \right) \right] \\

 & \Rightarrow {{\left( \cos A\sin B \right)}^{2}}+{{\left( \sin A\cos B \right)}^{2}}+2\left[ \left( \cos A \right)\left( \sin B \right) \right]\left[ \left( \sin A \right)\left( \cos B \right) \right] \\

\end{align}\]

We know the algebra identity ${{a}^{2}}+{{b}^{2}}+2ab={{\left( a+b \right)}^{2}}$. Now, by using this identity, we can further simplify our expression as

$\Rightarrow {{\left( \cos A\sin B+\sin A\cos B \right)}^{2}}$

Now, by using the identity sin (A + B) = sin A cos B + cos A sin B, we can rewrite the above expression as

$\Rightarrow {{\sin }^{2}}\left( A+B \right)$

Hence, we proved the equivalence of both sides by considering the expression of the left side.

Note: Students must remember the trigonometric formulas associated with different functions. This problem can be alternatively solved by expanding both cos (A+B) and sin (A+B) simultaneously and simplifying both the sides individually.