Question

Simplify ${\cos ^2}\left( {\dfrac{\theta }{2}} \right) - {\sin ^2}\left( {\dfrac{\theta }{2}} \right)$.

Answer 1

Hint:Using some basic trigonometric identities like we can simplify the above expression.
$\cos \left( {\dfrac{\theta }{2}} \right) = \sqrt {\dfrac{{1 + \cos \theta }}{2}} \\
\Rightarrow\sin \left( {\dfrac{\theta }{2}} \right) = \sqrt {\dfrac{{1 - \cos \theta }}{2}} \\ $
Such that in order to solve and simplify the given expression we have to use the above identities and express our given expression in that form and thereby simplify it.

Complete step by step answer:
Given, ${\cos ^2}\left( {\dfrac{\theta }{2}} \right) - {\sin ^2}\left( {\dfrac{\theta }{2}} \right)........................\left( i \right)$
Now to simplify this expression we have to use some basic trigonometric identities. We have to convert the given equation in the form of identities known to us and thereby simplify it.So we know that
$\cos \left( {\dfrac{\theta }{2}} \right) = \sqrt {\dfrac{{1 + \cos \theta }}{2}} \\
\Rightarrow \sin \left( {\dfrac{\theta }{2}} \right) = \sqrt {\dfrac{{1 - \cos \theta }}{2}} \\ $
Now we have to substitute these basic identities to the given equation (i) and then further solve:Such that we can write:
${\cos ^2}\left( {\dfrac{\theta }{2}} \right) = {\left( {\sqrt {\dfrac{{1 + \cos \theta }}{2}} } \right)^2}.......................\left( {ii} \right) \\
\Rightarrow {\sin ^2}\left( {\dfrac{\theta }{2}} \right) = {\left( {\sqrt {\dfrac{{1 - \cos \theta }}{2}} } \right)^2}.......................\left( {iii} \right) \\ $
So let’s substitute (ii) and (iii) in (i). Now we can write:
${\cos ^2}\left( {\dfrac{\theta }{2}} \right) - {\sin ^2}\left( {\dfrac{\theta }{2}} \right) = {\left( {\sqrt {\dfrac{{1 + \cos \theta }}{2}} } \right)^2} - {\left( {\sqrt {\dfrac{{1 - \cos \theta }}{2}} } \right)^2}..............\left( {iv} \right)$
Now we just have to expand the bracket in RHS and solve the equation (iv) using simple mathematical techniques. Such that by simply simplifying the above equation we can find the answer.So equation (iv) becomes:
${\cos ^2}\left( {\dfrac{\theta }{2}} \right) - {\sin ^2}\left( {\dfrac{\theta }{2}} \right) = {\left( {\sqrt {\dfrac{{1 + \cos \theta }}{2}} } \right)^2} - {\left( {\sqrt {\dfrac{{1 - \cos \theta }}{2}} } \right)^2} \\
\Rightarrow{\cos ^2}\left( {\dfrac{\theta }{2}} \right) - {\sin ^2}\left( {\dfrac{\theta }{2}} \right) = \dfrac{{1 + \cos \theta }}{2} - \dfrac{{1 - \cos \theta }}{2}......................\left( v \right) \\ $
Now since in equation (v) the terms in RHS have the same denominator we can write:
${\cos ^2}\left( {\dfrac{\theta }{2}} \right) - {\sin ^2}\left( {\dfrac{\theta }{2}} \right) = \dfrac{{1 + \cos \theta }}{2} - \dfrac{{1 - \cos \theta }}{2} \\
\Rightarrow {\cos ^2}\left( {\dfrac{\theta }{2}} \right) - {\sin ^2}\left( {\dfrac{\theta }{2}} \right) = \dfrac{{1 + \cos \theta - \left( {1 - \cos \theta } \right)}}{2} \\
\Rightarrow {\cos ^2}\left( {\dfrac{\theta }{2}} \right) - {\sin ^2}\left( {\dfrac{\theta }{2}} \right) = \dfrac{{1 + \cos \theta - 1 + \cos \theta }}{2} \\
\Rightarrow {\cos ^2}\left( {\dfrac{\theta }{2}} \right) - {\sin ^2}\left( {\dfrac{\theta }{2}} \right) = \dfrac{{2\cos \theta }}{2} \\
\therefore{\cos ^2}\left( {\dfrac{\theta }{2}} \right) - {\sin ^2}\left( {\dfrac{\theta }{2}} \right) = \cos \theta ......................\left( {vi} \right) \\ $
Therefore on simplifying ${\cos ^2}\left( {\dfrac{\theta }{2}} \right) - {\sin ^2}\left( {\dfrac{\theta }{2}} \right)$ we get: $\cos \theta $.

Note:Some other equations needed for solving these types of problem are:
$\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }} \\
\Rightarrow 2{\cos ^2}\theta - 1 = \cos 2\theta \\
\Rightarrow 1 + {\tan ^2}\theta = {\sec ^2}\theta $
Also while approaching a trigonometric problem one should keep in mind that one should work with one side at a time and manipulate it to the other side. The most straightforward way to do this is to simplify one side to the other directly, but we can also transform both sides to a common expression if we see no direct way to connect the two.