Question

How do you prove the trigonometric equation \[{\left( {\cos A + \cos B} \right)^2} + {\left( {\sin A + \sin B} \right)^2} = 2\left[ {1 + \cos \left( {A - B} \right)} \right]\].

Answer 1

Hint: Simplify the LHS (Left Hand Side) of the equation with the use of trigonometric and algebraic identities and obtain the RHS (Right Hand Side).
The trigonometric identities that are used to prove this equation are shown below.
\[{\sin ^2}\theta + {\cos ^2}\theta = 1\] --- (1)
\[\cos \theta \cos \phi + \sin \theta \sin \phi = \cos \left( {\theta - \phi } \right)\] ------ (2)

The algebraic identity that is used to expand the left hand side terms is written below.
\[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\] ---- (3)

Complete step by step solution:
Take Left hand side of the given equation and expand both terms by the use of algebraic identity \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\] as shown below.
\[LHS = {\left( {\cos A + \cos B} \right)^2} + {\left( {\sin A + \sin B} \right)^2}\]
\[ \Rightarrow \left( {{{\cos }^2}A + {{\cos }^2}B + 2 \cdot \cos A \cdot \cos B} \right) + \left( {{{\sin }^2}A + {{\sin }^2}B + 2 \cdot \sin A \cdot \sin B} \right)\]

Rearrange terms from the above expression as shown below.
\[ \Rightarrow \left( {{{\sin }^2}A + {{\cos }^2}A} \right) + \left( {{{\sin }^2}B + {{\cos }^2}B} \right) + \left( {2 \cdot \cos A \cdot \cos B + 2 \cdot \sin A \cdot \sin B} \right)\]

Use the trigonometry identity \[{\sin ^2}\theta + {\cos ^2}\theta = 1\] and simplify as shown below.
\[ \Rightarrow \left( 1 \right) + \left( 1 \right) + \left( {2 \cdot \cos A \cdot \cos B + 2 \cdot \sin A \cdot \sin B} \right)\]
\[ \Rightarrow 2 + 2\left( {\cos A \cdot \cos B + \sin A \cdot \sin B} \right)\]

Now, use the trigonometric identity \[\cos \theta \cos \phi + \sin \theta \sin \phi = \cos \left( {\theta - \phi } \right)\] and simplify the result as follows:
\[ \Rightarrow 2 + 2\cos \left( {A - B} \right)\]
\[ \Rightarrow 2\left[ {1 + \cos \left( {A - B} \right)} \right] = RHS\]

Hence, the trigonometric equation \[{\left( {\cos A + \cos B} \right)^2} + {\left( {\sin A + \sin B} \right)^2} = 2\left[ {1 + \cos \left( {A - B} \right)} \right]\] is proved.

Note: Always solve these types of questions very carefully as there are very high chances to go wrong in trigonometric questions. Memorize all the basic values of trigonometric function at standard angles as shown in table 1 and understand all the conversion formulas very carefully.
Some basic values of trigonometric functions at standard angles are shown below in table 1

	\(0^\circ \)	\(30^\circ \)	\(45^\circ \)	\(60^\circ \)	\(90^\circ \)
\(\sin \)	\(0\)	\(\dfrac{1}{2}\)	\(\dfrac{1}{{\sqrt 2 }}\)	\(\dfrac{{\sqrt 3 }}{2}\)	\(1\)
\(\cos \)	\(1\)	\(\dfrac{{\sqrt 3 }}{2}\)	\(\dfrac{1}{{\sqrt 2 }}\)	\(\dfrac{1}{2}\)	\(0\)
\(\tan \)	\(0\)	\(\dfrac{1}{{\sqrt 3 }}\)	\(1\)	\(\sqrt 3 \)	Not defined
Table 1