Question

Prove that: \[{\left( {\cos x + \cos y} \right)^2} + {\left( {\sin x - \sin y} \right)^2} = 4{\cos ^2}\left( {\dfrac{{x + y}}{2}} \right)\].

Answer 1

Hint:
Here, we need to prove the given equation. We will simplify the expression on the left hand side using algebraic identities. Then, we will use trigonometric identities to simplify the expression such that it is equal to the right hand side, and hence, prove the given equation.
Formula used: We will use the following formulas:
1) The square of the sum of two numbers is given by the algebraic identity \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\].
2) The square of the difference of two numbers is given by the algebraic identity \[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\].
3) The sum of the square of sine and cosine of an angle \[\theta \] is equal to 1, that is \[{\sin ^2}\theta + {\cos ^2}\theta = 1\].
4) The cosine of sum of two angles \[A\] and \[B\] is given by \[\cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B\].
5) The cosine of a double angle is given by the formula \[\cos 2A = 2{\cos ^2}A - 1\].

Complete step by step solution:
We will use algebraic identities and trigonometric identities to simplify the expression \[{\left( {\cos x + \cos y} \right)^2} + {\left( {\sin x - \sin y} \right)^2}\].
The square of the sum of two numbers is given by the algebraic identity \[{\left( {a + b} \right)^2} = {a^2} + {b^2} + 2ab\].
Substituting \[a = \cos x\] and \[b = \cos y\] in the algebraic identity, we get
\[ \Rightarrow {\left( {\cos x + \cos y} \right)^2} = {\left( {\cos x} \right)^2} + {\left( {\cos y} \right)^2} + 2\left( {\cos x} \right)\left( {\cos y} \right)\]
Rewriting the equation, we get
\[ \Rightarrow {\left( {\cos x + \cos y} \right)^2} = {\cos ^2}x + {\cos ^2}y + 2\cos x\cos y\]
The square of the difference of two numbers is given by the algebraic identity \[{\left( {a - b} \right)^2} = {a^2} + {b^2} - 2ab\].
Substituting \[a = \sin x\] and \[b = \sin y\] in the algebraic identity, we get
\[ \Rightarrow {\left( {\sin x - \sin y} \right)^2} = {\left( {\sin x} \right)^2} + {\left( {\sin y} \right)^2} - 2\left( {\sin x} \right)\left( {\sin y} \right)\]
Rewriting the equation, we get
\[ \Rightarrow {\left( {\sin x - \sin y} \right)^2} = {\sin ^2}x + {\sin ^2}y - 2\sin x\sin y\]
Now, substituting \[{\left( {\cos x + \cos y} \right)^2} = {\cos ^2}x + {\cos ^2}y + 2\cos x\cos y\] and \[{\left( {\sin x - \sin y} \right)^2} = {\sin ^2}x + {\sin ^2}y - 2\sin x\sin y\] in the expression \[{\left( {\cos x + \cos y} \right)^2} + {\left( {\sin x - \sin y} \right)^2}\], we get
\[{\left( {\cos x + \cos y} \right)^2} + {\left( {\sin x - \sin y} \right)^2} = {\cos ^2}x + {\cos ^2}y + 2\cos x\cos y + {\sin ^2}x + {\sin ^2}y - 2\sin x\sin y\]
Rewriting the equation using parentheses, we get
\[ \Rightarrow {\left( {\cos x + \cos y} \right)^2} + {\left( {\sin x - \sin y} \right)^2} = \left( {{{\sin }^2}x + {{\cos }^2}x} \right) + \left( {{{\sin }^2}y + {{\cos }^2}y} \right) + 2\left( {\cos x\cos y - \sin x\sin y} \right)\]
Now, we know that the sum of the square of sine and cosine of an angle \[\theta \] is equal to 1, that is \[{\sin ^2}\theta + {\cos ^2}\theta = 1\].
Therefore, we get
\[{\sin ^2}x + {\cos ^2}x = 1\] and \[{\sin ^2}y + {\cos ^2}y = 1\]
Substituting \[{\sin ^2}x + {\cos ^2}x = 1\] and \[{\sin ^2}y + {\cos ^2}y = 1\] in the equation, we get
\[ \Rightarrow {\left( {\cos x + \cos y} \right)^2} + {\left( {\sin x - \sin y} \right)^2} = 1 + 1 + 2\left( {\cos x\cos y - \sin x\sin y} \right)\]
Adding the terms, we get
\[ \Rightarrow {\left( {\cos x + \cos y} \right)^2} + {\left( {\sin x - \sin y} \right)^2} = 2 + 2\left( {\cos x\cos y - \sin x\sin y} \right)\]
The cosine of sum of two angles \[A\] and \[B\] is given by \[\cos \left( {A + B} \right) = \cos A\cos B - \sin A\sin B\].
Substituting \[A = x\] and \[B = y\] in the formula, we get
\[\cos \left( {x + y} \right) = \cos x\cos y - \sin x\sin y\]
Substituting \[\cos x\cos y - \sin x\sin y = \cos \left( {x + y} \right)\] in the equation \[{\left( {\cos x + \cos y} \right)^2} + {\left( {\sin x - \sin y} \right)^2} = 2 + 2\left( {\cos x\cos y - \sin x\sin y} \right)\], we get
\[ \Rightarrow {\left( {\cos x + \cos y} \right)^2} + {\left( {\sin x - \sin y} \right)^2} = 2 + 2\cos \left( {x + y} \right)\]
Factoring out 2 from the expression, we get
\[ \Rightarrow {\left( {\cos x + \cos y} \right)^2} + {\left( {\sin x - \sin y} \right)^2} = 2\left[ {1 + \cos \left( {x + y} \right)} \right]\]
Now, we know that the cosine of a double angle is given by the formula \[\cos 2A = 2{\cos ^2}A - 1\].
Substituting \[A = \dfrac{{x + y}}{2}\] in the formula, we get
\[\cos \left[ {2\left( {\dfrac{{x + y}}{2}} \right)} \right] = 2{\cos ^2}\left( {\dfrac{{x + y}}{2}} \right) - 1\]
Simplifying the expression, we get
\[ \Rightarrow \cos \left( {x + y} \right) = 2{\cos ^2}\left( {\dfrac{{x + y}}{2}} \right) - 1\]
Substituting \[\cos \left( {x + y} \right) = 2{\cos ^2}\left( {\dfrac{{x + y}}{2}} \right) - 1\] in the equation \[{\left( {\cos x + \cos y} \right)^2} + {\left( {\sin x - \sin y} \right)^2} = 2\left[ {1 + \cos \left( {x + y} \right)} \right]\], we get
\[ \Rightarrow {\left( {\cos x + \cos y} \right)^2} + {\left( {\sin x - \sin y} \right)^2} = 2\left[ {1 + 2{{\cos }^2}\left( {\dfrac{{x + y}}{2}} \right) - 1} \right]\]
Subtracting 1 from 1 in the parentheses, we get
\[ \Rightarrow {\left( {\cos x + \cos y} \right)^2} + {\left( {\sin x - \sin y} \right)^2} = 2\left[ {2{{\cos }^2}\left( {\dfrac{{x + y}}{2}} \right)} \right]\]
Multiplying the terms 2 and \[2{\cos ^2}\left( {\dfrac{{x + y}}{2}} \right)\], we get
\[{\left( {\cos x + \cos y} \right)^2} + {\left( {\sin x - \sin y} \right)^2} = 4{\cos ^2}\left( {\dfrac{{x + y}}{2}} \right)\]

Therefore, we have proved that \[{\left( {\cos x + \cos y} \right)^2} + {\left( {\sin x - \sin y} \right)^2}\] is equal to \[4{\cos ^2}\left( {\dfrac{{x + y}}{2}} \right)\].

Note:
We need to know the basics of trigonometric identities to solve the question. We can make a mistake is to simplify the expression \[2{\cos ^2}\left( {\dfrac{{x + y}}{2}} \right)\] as \[{\cos ^2}\left( {x + y} \right)\]. This is incorrect as the 2 cannot be cancelled. The term in the parentheses is the measure of the angle whose cosine is taken, and unless a trigonometric identity is used, we cannot change the angle.