Question

Find out which one of the following statements is true for the expression \[\cos^{2}\left( {A - B} \right) + \cos^{2}B – 2\cos\left( {A - B} \right)\cos A \cos B\] .
A. Dependent on B only.
B. Dependent on A and B.
C. Dependent on A only.
D. Independent of A and B.

Answer 1

Hint: First, factor out the common terms from the given equation. Then simplify the equation using the trigonometric identity subtraction formula for cosine. Further, simplify the equation and get the term as \[\cos\left( {A + B} \right)\]. After that, simplify the equation by using the trigonometric identity product of cosine. In the end, apply the trigonometric identity \[\sin^{2}A + \cos^{2}A = 1\] and find the value of the given expression. Verify the variables present in the output and get the required answer.

Formula used
\[\cos\left( {A - B} \right) = \cos A \cos B + \sin A \sin B\]
\[\cos\left( {A + B} \right) = \cos A \cos B - \sin A \sin B\]
\[\cos\left( {A + B} \right)\cos\left( {A - B} \right) = \cos^{2} A - \sin^{2} B\]
\[\sin^{2}A + \cos^{2}A = 1\]

Complete step by step solution
The given expression is \[\cos^{2}\left( {A - B} \right) + \cos^{2}B – 2\cos\left( {A - B} \right)\cos A \cos B\].
Let \[V\] be the value of the given expression.
Then,
\[V = \cos^{2}B + \cos^{2}\left( {A - B} \right) – 2\cos\left( {A - B} \right)\cos A \cos B\]
Factor out the common term from the right-hand side.
\[V = \cos^{2}B + \cos\left( {A - B} \right)\left[ {\cos\left( {A - B} \right) – 2\cos A \cos B} \right]\]
Apply the formula of the difference of angles for cosine function \[\cos\left( {A - B} \right) = \cos A \cos B + \sin A \sin B\].
We get,
\[V = \cos^{2}B + \cos\left( {A - B} \right)\left[ {\cos A \cos B + \sin A \sin B – 2\cos A \cos B} \right]\]
\[ \Rightarrow V = \cos^{2}B + \cos\left( {A - B} \right)\left[ {\sin A \sin B - \cos A \cos B} \right]\]
\[ \Rightarrow V = \cos^{2}B - \cos\left( {A - B} \right)\left[ {\cos A \cos B - \sin A \sin B} \right]\]
Apply the identity \[\cos\left( {A + B} \right) = \cos A \cos B - \sin A \sin B\].
\[V = \cos^{2}B - \cos\left( {A - B} \right)\cos\left( {A + B} \right)\]
Now use the trigonometric formula \[\cos\left( {A + B} \right)\cos\left( {A - B} \right) = \cos^{2} A - \sin^{2} B\].
\[V = \cos^{2}B - \left( {\cos^{2} A - \sin^{2} B} \right)\]
\[ \Rightarrow V = \cos^{2}B - \cos^{2} A + \sin^{2} B\]
\[ \Rightarrow V = \sin^{2} B + \cos^{2}B - \cos^{2} A\]
Apply the trigonometric identity \[\sin^{2}A + \cos^{2}A = 1\].
\[V = 1 - \cos^{2} A\]
Again, apply the trigonometric identity \[\sin^{2}A + \cos^{2}A = 1\].
We get,
\[V = \sin^{2} A\]
Thus,
\[\cos^{2}\left( {A - B} \right) + \cos^{2}B – 2\cos\left( {A - B} \right)\cos A \cos B = \sin^{2} A\]
From the above equation, we observe that the value of the given expression is in terms of variable A.
Therefore, the given expression is dependent on A only.

Hence option C is correct.

Note: While solving any trigonometric equation with double angles and exponent convert the equation into the basic trigonometric ratios by using the trigonometric identities.
Students often get confused about the formulas of the sum and difference of angles for a cosine function.
The formulas are as follows:
\[\cos\left( {A - B} \right) = \cos A \cos B + \sin A \sin B\]
\[\cos\left( {A + B} \right) = \cos A \cos B - \sin A \sin B\]