What is ${\cos ^2}{48^ \circ } - {\sin ^2}{12^ \circ }$ equal to?
A. $\dfrac{{\sqrt 5 - 1}}{8}$
B. $\dfrac{{\sqrt 5 + 1}}{8}$
C. $\dfrac{{\sqrt 3 - 1}}{4}$
D. $\dfrac{{\sqrt 3 + 1}}{{2\sqrt 2 }}$

Hint:We need be aware that \[\sin \theta = \cos \left( {{{90}^ \circ } - \theta } \right)\] in order to demonstrate the relationship. Additionally, we should be familiar with a few trigonometric addition formulas, such as \[cosA + cosB = 2cos(\frac{{A + B}}{2})cos(\frac{{A - B}}{2})\] and \[cosA - cosB = - 2sin(\frac{{A + B}}{2})sin(\frac{{A - B}}{2})\] we also need to be aware that \[{a^2} - {b^2} = (a - b)(a + b)\]. These relationships allow us to demonstrate the necessary outcome.

Formula used:
\[cosA + cosB = 2cos(\frac{{A + B}}{2})cos(\frac{{A - B}}{2})\]

Complete step by step solution:
We will first look at the left-hand side, or LHS, of the desired expression to demonstrate this relationship.
\[{\cos ^2}{48^\circ } - {\sin ^2}{12^\circ }\]
Since we are aware that \[\sin \theta = \cos \left( {{{90}^ \circ } - \theta } \right)\] we can utilize that knowledge to get LHS as,
\[ = {\cos ^2}{48^ \circ } - {\cos ^2}(90 - {12^ \circ })\]
Now, let us simplify the terms inside the parentheses, we get
\[ = {\cos ^2}{48^ \circ } - {\cos ^2}({78^ \circ })\]
Now that we are aware that \[{a^2} - {b^2} = (a - b)(a + b)\] applies to the preceding statement, we may obtain the LHS as,
\[ = (\cos {48^ \circ } + \cos {78^ \circ })(\cos {48^ \circ } - \cos {78^ \circ })\]
We have been already aware that,
\[\cos A + \cos B = 2\cos \left( {\frac{{A + B}}{2}} \right)\cos \left( {\frac{{A - B}}{2}} \right)\]
And
\[\cos A - \cos B = - 2\sin \left( {\frac{{A + B}}{2}} \right)\sin \left( {\frac{{A - B}}{2}} \right)\]
Now, on applying the above-mentioned condition, we get the LHS as,
\[ = \left[ {2\cos \left( {\frac{{{{48}^\circ } + {{78}^\circ }}}{2}} \right)\cos \left( {\frac{{{{48}^\circ } - 78}}{2}^\circ } \right)} \right] \times \left[ { - 2\sin \left( {\frac{{{{48}^\circ } + {{78}^\circ }}}{2}} \right)\sin \left( {\frac{{{{48}^\circ } - {{78}^\circ }}}{2}} \right)} \right]\]
Now, we have to simplify the terms inside the parentheses, we get
\[ = \left[ {2\left( {\cos {{63}^\circ }} \right)\left( {\cos {{15}^\circ }} \right)} \right] \times \left[ { - 2\left( {\sin {{63}^\circ }} \right)\left( {\sin \left( { - {{15}^\circ }} \right)} \right)} \right]\]
Now, let us rearrange the terms from the above equation, we get
\[ = [2(\sin {63^ \circ })(\cos {63^ \circ })] \times [ - 2\sin ( - {15^ \circ })\cos ({15^ \circ })]\]
Now, we have been already known the condition that,
\[\sin \left( { - \theta } \right) = - \sin \theta \]
And
\[2\sin \theta \cos \theta = \sin 2\theta \]
Now, on applying the above-mentioned condition, we get the LHS as
\[ = \left[ {2\left( {\sin {{63}^\circ }} \right)\left( {\cos {{63}^\circ }} \right)} \right] \times \left[ {2\left( {\sin {{15}^\circ }} \right)\left( {\cos {{15}^\circ }} \right)} \right]\]
Now, on further simplification of the above equation, we get
\[ = \sin 2\left( {{{63}^\circ }} \right) \times \sin 2\left( {{{15}^\circ }} \right)\]
Now, multiply the terms, we get
\[ = \sin \left( {{{126}^\circ }} \right) \times \sin \left( {{{30}^\circ }} \right)\]
Now, we have been already known the condition that,
\[\sin {30^ \circ } = \frac{1}{2}\]
And
\[\sin \left( {90 + \theta } \right) = \cos \theta \]
Now, on applying the above-mentioned condition and rearranging it, we get
\[ = \sin {30^ \circ }\sin ({90^ \circ } + {36^ \circ })\]
\[ = \frac{1}{2}\cos {36^ \circ }\]
We ought to be aware that,
\[\cos {36^ \circ } = \frac{{\sqrt 5 + 1}}{4}\]
In the above, by substituting the same, we obtain the LHS as,
\[ = \frac{1}{2} \cdot \frac{{\sqrt 5 + 1}}{4}\]
On further simplification, we get
\[ = \frac{{\sqrt 5 + 1}}{8}\]
The phrase given on the right side of the expression in the proof-needed question is the same as this.
Therefore, \[{\cos ^2}{48^\circ } - {\sin ^2}{12^\circ }\]is equal to \[\frac{{\sqrt 5 + 1}}{8}\]
Hence, the option B is correct

Note: Students often make mistakes in these types of problems because it includes trigonometry functions. So, one should keep in mind that applying trig formulas correctly will give the desired result. It is also important to remember that by assuming that \[\cos {36^ \circ }\] has a value of \[{18^ \circ } = A\] and a value of \[{90^ \circ } = 5A\], we can also determine its value.