How do you prove that $1 + {\cos ^2}\left( {\dfrac{x}{2}} \right) = \dfrac{{{{\sin }^2}x}}{{2(1 + \cos x)}}$ ?
Answer
615.3k+ views
Hint: By using the basic trigonometric identity given below we can prove the above expression that is $1 + {\cos ^2}\left( {\dfrac{x}{2}} \right) = \dfrac{{{{\sin }^2}x}}{{2(1 + \cos x)}}$ . We can prove the given statement by using ${\sin ^2}x = 1 - {\cos ^2}x$ and $\cos x = 2{\cos ^2}\left( {\dfrac{x}{2}} \right) - 1$ . In order to solve and simplify the given expression we have to use the identity and express our given expression in the simplest form and thereby solve it.
Complete step-by-step solution:
To prove: $1 + {\cos ^2}\left( {\dfrac{x}{2}} \right) = \dfrac{{{{\sin }^2}x}}{{2(1 + \cos x)}}$
Proof:
Taking R.H.S of the above expression ,
$ \Rightarrow \dfrac{{{{\sin }^2}x}}{{2(1 + \cos x)}}$
We already know that ${\sin ^2}x = 1 - {\cos ^2}x$ ,
Therefore, we can write ,
$ \Rightarrow \dfrac{{1 - {{\cos }^2}x}}{{2(1 + \cos x)}}$
Now using the identity $({a^2} - {b^2}) = (a - b)(a + b)$,
Therefore, we can write as ,
$ \Rightarrow \dfrac{{(1 + \cos x)(1 - \cos x)}}{{2(1 + \cos x)}}$
Now simplify the above equation,
$ \Rightarrow \dfrac{{(1 - \cos x)}}{2}$
Now, using the trigonometric identity $\cos x = 2{\cos ^2}\left( {\dfrac{x}{2}} \right) - 1$ ,
$ \Rightarrow \dfrac{{\left( {1 - \left( {2{{\cos }^2}\left( {\dfrac{x}{2}} \right) - 1} \right)} \right)}}{2}$
Now simplify the above equation,
$
\Rightarrow \dfrac{{\left( {1 - 2{{\cos }^2}\left( {\dfrac{x}{2}} \right) + 1} \right)}}{2} \\
\Rightarrow \dfrac{{2 + 2{{\cos }^2}\left( {\dfrac{x}{2}} \right)}}{2} \\
\Rightarrow 1 + {\cos ^2}\left( {\dfrac{x}{2}} \right) \\
$
It is equal to L.H.S. of the given problem ,
Since L.H.S. = R.H.S.
Hence proved LHS = RHS.
Note: Some other equations needed for solving these types of problem are:
${\sin ^2}x = 1 - {\cos ^2}x$ and $\cos x = 2{\cos ^2}\left( {\dfrac{x}{2}} \right) - 1$ .
Range of cosine and sine is $\left[ { - 1,1} \right]$ ,
Range of tangents is minus infinity to infinity .
In order to solve and simplify the given expression we have to use the identity and express our given expression in the simplest form and thereby solve it. 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.
Complete step-by-step solution:
To prove: $1 + {\cos ^2}\left( {\dfrac{x}{2}} \right) = \dfrac{{{{\sin }^2}x}}{{2(1 + \cos x)}}$
Proof:
Taking R.H.S of the above expression ,
$ \Rightarrow \dfrac{{{{\sin }^2}x}}{{2(1 + \cos x)}}$
We already know that ${\sin ^2}x = 1 - {\cos ^2}x$ ,
Therefore, we can write ,
$ \Rightarrow \dfrac{{1 - {{\cos }^2}x}}{{2(1 + \cos x)}}$
Now using the identity $({a^2} - {b^2}) = (a - b)(a + b)$,
Therefore, we can write as ,
$ \Rightarrow \dfrac{{(1 + \cos x)(1 - \cos x)}}{{2(1 + \cos x)}}$
Now simplify the above equation,
$ \Rightarrow \dfrac{{(1 - \cos x)}}{2}$
Now, using the trigonometric identity $\cos x = 2{\cos ^2}\left( {\dfrac{x}{2}} \right) - 1$ ,
$ \Rightarrow \dfrac{{\left( {1 - \left( {2{{\cos }^2}\left( {\dfrac{x}{2}} \right) - 1} \right)} \right)}}{2}$
Now simplify the above equation,
$
\Rightarrow \dfrac{{\left( {1 - 2{{\cos }^2}\left( {\dfrac{x}{2}} \right) + 1} \right)}}{2} \\
\Rightarrow \dfrac{{2 + 2{{\cos }^2}\left( {\dfrac{x}{2}} \right)}}{2} \\
\Rightarrow 1 + {\cos ^2}\left( {\dfrac{x}{2}} \right) \\
$
It is equal to L.H.S. of the given problem ,
Since L.H.S. = R.H.S.
Hence proved LHS = RHS.
Note: Some other equations needed for solving these types of problem are:
${\sin ^2}x = 1 - {\cos ^2}x$ and $\cos x = 2{\cos ^2}\left( {\dfrac{x}{2}} \right) - 1$ .
Range of cosine and sine is $\left[ { - 1,1} \right]$ ,
Range of tangents is minus infinity to infinity .
In order to solve and simplify the given expression we have to use the identity and express our given expression in the simplest form and thereby solve it. 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.
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