Question

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

Answer 1

Hint: The given question deals with basic simplification of trigonometric functions by using some of the simple trigonometric formulae such as $ \cos \left( {x + y} \right) = \cos x\cos y - \sin x\sin y $ and $ \cos \left( {x - y} \right) = \cos x\cos y + \sin x\sin y $ . Basic algebraic rules and trigonometric identities are to be kept in mind while doing simplification in the given problem and proving the result given to us.

Complete step by step solution:
In the given problem, we have to prove a trigonometric equality that can be further used in many questions and problems as a direct result and has wide ranging applications. For proving the desired result, we need to have a good grip over the basic trigonometric formulae and identities.
Now, we need to make the left and right sides of the equation equal.
L.H.S. $ = \cos \left( {A + B} \right)\cos \left( {A - B} \right) $
Now, we have to apply the compound angle formula in the expression given to us in the left hand side of the equation. Now, we know that $ \cos \left( {x + y} \right) = \cos x\cos y - \sin x\sin y $ and $ \cos \left( {x - y} \right) = \cos x\cos y + \sin x\sin y $ .
So, we get,
 $ \Rightarrow \left[ {\cos \left( A \right)\cos \left( B \right) - \sin \left( A \right)\sin \left( B \right)} \right]\left[ {\cos \left( A \right)\cos \left( B \right) + \sin \left( A \right)\sin \left( B \right)} \right] $
Using the algebraic identity $ \left( {a - b} \right)\left( {a + b} \right) = {a^2} - {b^2} $ , we get,
\[ \Rightarrow {\left[ {\cos \left( A \right)\cos \left( B \right)} \right]^2} - {\left[ {\sin \left( A \right)\sin \left( B \right)} \right]^2}\]
Simplifying the expressions, we get,
\[ \Rightarrow {\cos ^2}\left( A \right){\cos ^2}\left( B \right) - {\sin ^2}\left( A \right){\sin ^2}\left( B \right)\]
Now, using the trigonometric identity \[1 - {\cos ^2}\left( x \right) = {\sin ^2}\left( x \right)\], we get,
\[ \Rightarrow {\cos ^2}\left( A \right){\cos ^2}\left( B \right) - \left( {1 - {{\cos }^2}\left( A \right)} \right)\left( {1 - {{\cos }^2}\left( B \right)} \right)\]
Opening the brackets, we get,
\[ \Rightarrow {\cos ^2}\left( A \right){\cos ^2}\left( B \right) - \left[ {1 - {{\cos }^2}\left( A \right) - {{\cos }^2}\left( B \right) + {{\cos }^2}\left( A \right){{\cos }^2}\left( B \right)} \right]\]
\[ \Rightarrow {\cos ^2}\left( A \right){\cos ^2}\left( B \right) - 1 + {\cos ^2}\left( A \right) + {\cos ^2}\left( B \right) - {\cos ^2}\left( A \right){\cos ^2}\left( B \right)\]
Cancelling the like terms with opposite signs, we get,
\[ \Rightarrow {\cos ^2}\left( A \right) + {\cos ^2}\left( B \right) - 1\]
\[ \Rightarrow {\cos ^2}\left( A \right) - \left( {1 - {{\cos }^2}\left( B \right)} \right)\]
Again using the trigonometric identity \[1 - {\cos ^2}\left( x \right) = {\sin ^2}\left( x \right)\], we get,
\[ \Rightarrow {\cos ^2}\left( A \right) - {\sin ^2}\left( B \right)\]
Now, L.H.S\[ = {\cos ^2}\left( A \right) - {\sin ^2}\left( B \right)\]
R.H.S\[ = {\cos ^2}\left( A \right) - {\sin ^2}\left( B \right)\]
As the left side of the equation is equal to the right side of the equation, we have,
\[\cos \left( {A + B} \right)\cos \left( {A - B} \right) = {\cos ^2}A - {\sin ^2}B\]
Hence, Proved.

Note: Given problem deals with Trigonometric functions. For solving such problems, trigonometric formulae should be remembered by heart. Besides these simple trigonometric formulae, trigonometric identities are also of significant use in such type of questions where we have to simplify trigonometric expressions with help of basic knowledge of algebraic rules and operations.