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Question: Please prove that\[\cos \left( {a + b} \right)\cos \left( {a - b} \right) = {\cos ^2}a + {\cos ^2}b ...

Please prove thatcos⁡(a+b)cos⁡(a−b)=cos⁡2a+cos⁡2b−1\cos \left( {a + b} \right)\cos \left( {a - b} \right) = {\cos ^2}a + {\cos ^2}b - 1.

Explanation

Solution

To solve this question, we will start with the LHS and try to get the RHS from it. We will first expand the LHS by using the formula of cos⁡(a+b),cos⁡(a−b)\cos \left( {a + b} \right),\cos \left( {a - b} \right). Then we will further multiply the terms using the formula for (a+b)(a−b)\left( {a + b} \right)\left( {a - b} \right). Then we will solve further by taking out the common terms and using the trigonometric formula.

Formula used:
(a+b)(a−b)=a2−b2\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}
cos⁡(a+b)=cos⁡acos⁡b−sin⁡asin⁡b\cos \left( {a + b} \right) = \cos a\cos b - \sin a\sin b
cos⁡(a−b)=cos⁡acos⁡b+sin⁡asin⁡b\cos \left( {a - b} \right) = \cos a\cos b + \sin a\sin b
cos⁡2a+sin⁡2a=1{\cos ^2}a + {\sin ^2}a = 1

Complete step by step answer:
We have the LHS as:
cos⁡(a+b)cos⁡(a−b)\cos \left( {a + b} \right)\cos \left( {a - b} \right)
Now we will expand this expression using the formula for cos⁡(a+b),cos⁡(a−b)\cos \left( {a + b} \right),\cos \left( {a - b} \right). So, we get;
=(cos⁡acos⁡b−sin⁡asin⁡b)(cos⁡acos⁡b+sin⁡asin⁡b)= \left( {\cos a\cos b - \sin a\sin b} \right)\left( {\cos a\cos b + \sin a\sin b} \right)
Now, we can see that this is in the form of (a+b)(a−b)\left( {a + b} \right)\left( {a - b} \right). So, using the formula (a+b)(a−b)=a2−b2\left( {a + b} \right)\left( {a - b} \right) = {a^2} - {b^2}, we get;
=cos⁡2acos⁡2b−sin⁡2asin⁡2b= {\cos ^2}a{\cos ^2}b - {\sin ^2}a{\sin ^2}b
Now we will solve it using the formulacos⁡2a+sin⁡2a=1{\cos ^2}a + {\sin ^2}a = 1. So, we get;
=(1−sin⁡2a)cos⁡2b−sin⁡2a(1−cos⁡2b)= \left( {1 - {{\sin }^2}a} \right){\cos ^2}b - {\sin ^2}a\left( {1 - {{\cos }^2}b} \right)
Now we will expand the brackets. So, we have;
=cos⁡2b−cos⁡2bsin⁡2a−sin⁡2a+cos⁡2bsin⁡2a= {\cos ^2}b - {\cos ^2}b{\sin ^2}a - {\sin ^2}a + {\cos ^2}b{\sin ^2}a
Cancelling the terms, we get;
=cos⁡2b−sin⁡2a= {\cos ^2}b - {\sin ^2}a
Using the formula 1−cos⁡2a=sin⁡2a1 - {\cos ^2}a = {\sin ^2}a. We get;
=cos⁡2b−(1−cos⁡2a)= {\cos ^2}b - \left( {1 - {{\cos }^2}a} \right)
Expanding we get;
=cos⁡2b+cos⁡2a−1= {\cos ^2}b + {\cos ^2}a - 1
And this is the RHS.

Note:
One thing to note here is that in the second last step we have replaced the sine term using the formula and not the cosine term because in the RHS we have the cosine terms. So, the point is we have to keep in mind what we need to get finally and then solve the question. One can also start with the RHS and get the LHS.