How do you prove $\sin \left( {\dfrac{\pi }{2} - x} \right) = \sin \left( {\dfrac{\pi }{2} + x} \right)?$

Hint:In this question, we are going to prove that the two sides of the trigonometric expression are equal.
By using the sum and difference formula to the trigonometric expression we can get the required result.
Hence we can prove that both the trigonometric expressions are equal.
Formula used: The sum and difference formula can be expressed as
 $\begin{array}{*{20}{c}}
  {\sin \left( {\alpha \pm \beta } \right) = \sin \alpha \cos \beta \pm \cos \alpha \sin \beta } \\
  {}
\end{array}$

Complete step by step answer:In this question we are going to prove both the trigonometric expression are same.
First, write the given equation and mark it as $\left( 1 \right)$
 $\sin \left( {\dfrac{\pi }{2} - x} \right) = \sin \left( {\dfrac{\pi }{2} + x} \right)$ $\left( 1 \right)$
Now apply the difference identity to the left hand side in equation $\left( 1 \right)$we get,
 $\sin \left( {\dfrac{\pi }{2} - x} \right) = \sin \dfrac{\pi }{2}\cos x - \cos \dfrac{\pi }{2}\sin x$ $\left( 2 \right)$
Putting the value,
$\sin \left( {\dfrac{\pi }{2} - x} \right) = \left( 1 \right)\cos x - \left( 0 \right)\sin x$ $\left[ {\because \sin \dfrac{\pi }{2} = 1,\cos \dfrac{\pi }{2} = 0} \right]$
Simplify the terms,
$\sin \left( {\dfrac{\pi }{2} - x} \right) = \cos x - 0$
On Subtracting,
$\sin \left( {\dfrac{\pi }{2} - x} \right) = \cos x$ $\left( 2 \right)$
Next apply the sum identity to the right hand side in equation $\left( 1 \right)$we get,
$\sin \left( {\dfrac{\pi }{2} + x} \right) = \sin \dfrac{\pi }{2}\cos x + \cos \dfrac{\pi }{2}\sin x$
Putting the value ,
$\sin \left( {\dfrac{\pi }{2} + x} \right) = \left( 1 \right)\cos x + \left( 0 \right)\sin x$ $\left[ {\because \sin \dfrac{\pi }{2} = 1,\cos \dfrac{\pi }{2} = 0} \right]$
On simplify,
$\sin \left( {\dfrac{\pi }{2} + x} \right) = \cos x + 0$
On adding the terms,
$\sin \left( {\dfrac{\pi }{2} + x} \right) = \cos x$ $\left( 3 \right)$
Now substitute equation $\left( 2 \right)$and $\left( 3 \right)$in equation$\left( 1 \right)$we get,
$\sin \left( {\dfrac{\pi }{2} + x} \right) = \cos x = \sin \left( {\dfrac{\pi }{2} - x} \right)$
Therefore, both sides of the trigonometric expression are the same.
 We have proved that $\sin \left( {\dfrac{\pi }{2} - x} \right) = \sin \left( {\dfrac{\pi }{2} + x} \right) = \cos x$
Both the trigonometric expressions are equal to $\cos x$ .

Note:
We can also solve this problem by using another method as follows:
By using the formula,
$\sin x = - \sin \left( {x - \pi } \right)$
 $ - \sin \left( { - x} \right) = - \sin \left( x \right)$
Applying that formula to the given trigonometric expression we get,
 $ \Rightarrow \sin \left( {\dfrac{\pi }{2} - x} \right) = - \sin \left( {\left( {\dfrac{\pi }{2} - x} \right) - \pi } \right)$
We just open the brackets
 $ = - \sin \left( {\dfrac{\pi }{2} - \dfrac{{2\pi }}{2} - x} \right)$
Simplify the bracket,
 $ = - \sin \left( { - x - \dfrac{\pi }{2}} \right)$
Taking $( - )ve$Sign common,
 $ = - \sin \left( { - \left( {\dfrac{\pi }{2} + x} \right)} \right)$
Taking negative sign outside,
 $ = - \left( { - \sin \left( {\dfrac{\pi }{2} + x} \right)} \right)$
Now we can say,
 $\sin \left( {\dfrac{\pi }{2} - x} \right) = \sin \left( {\dfrac{\pi }{2} + x} \right)$
Hence we get the required result.

We can prove the identity by reducing each side separately. We do not know if the two sides are equal. At the end we ended up with the same thing, so we can conclude that it is valid.
Instead of equating the left hand side with the right hand side, subtract them.
There are several options we can use when proving a trigonometric identity.
Often one of the steps for proving identities to change each term into their sine and cosine equivalents.
Use Pythagorean Theorem and other fundamental identities.
When working with identities where there are fractions combine using algebraic techniques for adding expressions with unlike denominators.
If possible factor trigonometric expressions.