Question

The value of ${{\cos }^{2}}{{75}^{\circ }}+{{\cos }^{2}}{{45}^{\circ }}+{{\cos }^{2}}{{15}^{\circ }}-{{\cos }^{2}}{{30}^{\circ }}-{{\cos }^{2}}{{60}^{\circ }}$ is:
(a) $0$
(b) $1$
(c) $\dfrac{1}{2}$
(d) $\dfrac{1}{4}$

Answer 1

Hint: Convert ${{\cos }^{2}}{{15}^{\circ }}$ into ${{\sin }^{2}}{{75}^{\circ }}$ by using the complementary angle rule that is, $\cos \theta =\sin \left( {{90}^{\circ }}-\theta \right)$, then pair this with ${{\cos }^{2}}{{75}^{\circ }}$. Use ${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1$. Substitute the values of the remaining cosine of angles.

Complete step-by-step answer:
Since, we don’t know the value of ${{\cos }^{2}}{{75}^{\circ }}$and ${{\cos }^{2}}{{15}^{\circ }}$, so, we will use a different approach here and will change this ${{\cos }^{2}}{{15}^{\circ }}$ into ${{\sin }^{2}}{{75}^{\circ }}$by using the complementary angle formula. Two angles are complementary when their sum is ${{90}^{\circ }}$, that means, a right angle. For example: ${{23}^{\circ }}\text{ and }{{67}^{\circ }}$ are a pair of complementary angles. Complementary angles are always acute angles. Acute angles are the angles which are greater than ${{0}^{\circ }}$ and less than ${{90}^{\circ }}$. The angles which are greater than ${{90}^{\circ }}$ but less than ${{180}^{\circ }}$ are known as obtuse angles. There is one more term we may face while solving these types of problems, supplementary angles. Supplementary angles are the angles whose sum is ${{180}^{\circ }}$. For example: ${{83}^{\circ }}\text{ and }{{97}^{\circ }}$ can be considered as a pair of supplementary angles. Supplementary angle pairs will either be two right angles or be one acute angle and one obtuse angle. When two parallel lines are crossed by a third line, the same side interior angles will be supplementary.
Now, we come to the question. Here, we have to deal only with the concept of complementary angles. So, the given expression is:
${{\cos }^{2}}{{75}^{\circ }}+{{\cos }^{2}}{{45}^{\circ }}+{{\cos }^{2}}{{15}^{\circ }}-{{\cos }^{2}}{{30}^{\circ }}-{{\cos }^{2}}{{60}^{\circ }}$
Applying complementary angle rule: $\cos {{15}^{\circ }}=\sin \left( {{90}^{\circ }}-{{15}^{\circ }} \right)=\sin {{75}^{\circ }}$, we get, $\begin{aligned} & ={{\cos }^{2}}{{75}^{\circ }}+{{\cos }^{2}}{{45}^{\circ }}+{{\sin }^{2}}{{75}^{\circ }}+{{\cos }^{2}}{{30}^{\circ }}-{{\cos }^{2}}{{60}^{\circ }} \\\ & =({{\cos }^{2}}{{75}^{\circ }}+{{\sin }^{2}}{{75}^{\circ }})+{{\cos }^{2}}{{45}^{\circ }}-{{\cos }^{2}}{{30}^{\circ }}-{{\cos }^{2}}{{60}^{\circ }}.........................(i) \\\ \end{aligned}$
We know that, ${{\cos }^{2}}\theta +{{\sin }^{2}}\theta =1$, therefore equation (i) becomes,
$1+{{\cos }^{2}}{{45}^{\circ }}-{{\cos }^{2}}{{30}^{\circ }}-{{\cos }^{2}}{{60}^{\circ }}..................(ii)$
We know that, $\cos {{45}^{\circ }}=\dfrac{1}{\sqrt{2}},\cos {{30}^{\circ }}=\dfrac{\sqrt{3}}{2},\cos {{60}^{\circ }}=\dfrac{1}{2}$. Substituting these values in equation (ii), we get,
$\begin{aligned} & 1+{{\left( \dfrac{1}{\sqrt{2}} \right)}^{2}}-{{\left( \dfrac{\sqrt{3}}{2} \right)}^{2}}-{{\left( \dfrac{1}{2} \right)}^{2}} \\\ & =1+\dfrac{1}{2}-\dfrac{3}{4}-\dfrac{1}{4} \\\ & =\dfrac{3}{2}-1 \\\ & =\dfrac{1}{2} \\\ \end{aligned}$
Hence, option (c) is the correct answer.

Note: Here, we have to pair up those cosine angles whose values are not known to us. So, we paired ${{\cos }^{2}}{{75}^{\circ }}$ and ${{\cos }^{2}}{{15}^{\circ }}$ to simplify it. We can calculate the value of ${{\cos }^{2}}{{75}^{\circ }}$ and ${{\cos }^{2}}{{15}^{\circ }}$ but that will be a lengthy process. So, it is easier to apply the approach that we have used here.