By using the trigonometric formulas and identities prove the following equation,
\[\cos {{18}^{\circ }}-\sin {{18}^{\circ }}=\sqrt{2}\sin {{27}^{\circ }}\].

Hint: Convert the \[\cos \theta \] function to \[\sin \theta \] function or vice versa by using the complementary angles conversion formula, \[\cos \theta =\sin \left( 90-\theta \right)\] and then apply the transformation formula and then transform the sum or difference into product of the trigonometric functions. We will apply the formula of, \[\sin C-\sin D=2\cos \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{C-D}{2} \right)\] and then we will solve the question and reach our answer.

Complete step-by-step answer:
Now the first thing we need to do is to memorize the formulas of trigonometry which involves simply \[\sin \theta \] and \[\cos \theta \] functions being linked together with addition or subtraction from one side and multiplication from another side of the equal to sign. The formulas are applicable when there is either only \[\sin \theta \] function terms or only \[\cos \theta \] function terms linked with addition or subtraction on one side of the equal sign. There are total four formulas:-
\[\begin{align}
  & \sin C+\sin D=2\sin \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right) \\
 & \sin C-\sin D=2\cos \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{C-D}{2} \right) \\
 & \cos C+\cos D=2\cos \left( \dfrac{C+D}{2} \right)\cos \left( \dfrac{C-D}{2} \right) \\
 & \cos C-\cos D=-2\sin \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{C-D}{2} \right) \\
\end{align}\]
We are going to use the second formula from this list but in the question we have a \[\cos \theta \] term instead of a \[\sin \theta \] along with the other \[\sin \theta \] term. So we are going to convert the \[\cos \theta \] term to \[\sin \theta \] term by using complementary angles transformation formula like \[\cos \theta =\sin \left( 90-\theta \right)\].
Now the L.H.S. of the equation is,
\[L.H.S.=\cos {{18}^{\circ }}-\sin {{18}^{\circ }}\]
After converting the \[\cos \theta \] term to \[\sin \theta \] term by using complementary angles transformation formula like \[\cos \theta =\sin \left( 90-\theta \right)\], we get,
\[\begin{align}
  & L.H.S.=\sin {{\left( 90-18 \right)}^{\circ }}-\sin {{18}^{\circ }} \\
 & L.H.S.=\sin {{72}^{\circ }}-\sin {{18}^{\circ }} \\
\end{align}\]
Now we can apply the formula to transform the sum or difference into product, the formula used is \[\sin C-\sin D=2\cos \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{C-D}{2} \right)\], we get,
\[\begin{align}
  & \sin {{72}^{\circ }}-\sin {{18}^{\circ }}=2\cos \left( \dfrac{72+18}{2} \right)\sin \left( \dfrac{72-18}{2} \right) \\
 & \sin {{72}^{\circ }}-\sin {{18}^{\circ }}=2\cos \left( \dfrac{90}{2} \right)\sin \left( \dfrac{54}{2} \right) \\
 & \sin {{72}^{\circ }}-\sin {{18}^{\circ }}=2\cos {{45}^{\circ }}\sin {{27}^{\circ }} \\
\end{align}\]
\[\sin {{72}^{\circ }}-\sin {{18}^{\circ }}=2\dfrac{1}{\sqrt{2}}\sin {{27}^{\circ }}\]
Now \[2\] can be written as \[{{\left( \sqrt{2} \right)}^{2}}\] which can further be written as \[\sqrt{2}\times \sqrt{2}\], now we get,
\[\sin {{72}^{\circ }}-\sin {{18}^{\circ }}=\left( \sqrt{2}\times \sqrt{2} \right)\dfrac{1}{\sqrt{2}}\sin {{27}^{\circ }}\]
\[\sin {{72}^{\circ }}-\sin {{18}^{\circ }}=\sqrt{2}\sin {{27}^{\circ }}\]
Now the L.H.S. we have evaluated is equal to the Right Hand Side part of the equation given in the question,
Hence proved.

Note: You need to memorize all the formulas of trigonometry because without them you will not be able to solve the questions. Sometimes solving the Right Hand Side is easy, so always look out for which side is easy to solve and which side has formation to apply the formulas. The formula \[\cos C-\cos D=-2\sin \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{C-D}{2} \right)=2\sin \left( \dfrac{C+D}{2} \right)\sin \left( \dfrac{D-C}{2} \right)\], so do not get confused. Sometimes the angle might be greater than \[{{90}^{\circ }}\] and we might have to use another quadrant (Cartesian) complementary angle transformation formula. You can also go for conversion of \[\sin \theta \] function to \[\cos \theta \] function as an alternative method.