Question

Given $\cos {{12}^{o}}+\cos {{84}^{o}}+\cos {{156}^{o}}+\cos {{132}^{o}}=-\dfrac{1}{a}$. Find the value of a.

Answer 1

Hint: We should know about the common trigonometric identities and numeric values to solve this problem.

Complete step-by-step answer:
To solve the above question, we can see that there are four terms containing cosine. Thus, to solve, we will group these four terms in pairs of two and then use the below property on each of the pairs to solve the problem-
$\cos A+\cos B=\cos \left( \dfrac{A+B}{2} \right)\cos \left( \dfrac{A-B}{2} \right)$
Further, we should also know the values to following trigonometric angles-
$\sin $${{18}^{o}}$ = $\dfrac{\sqrt{5}-1}{4}$
$\cos $${{36}^{o}}$= $\dfrac{\sqrt{5}+1}{4}$

Now, to begin the question, we need to decide which terms to club together to solve the problem efficiently. To explain this,
We can club ($\cos $${{12}^{o}}$ and $\cos $${{84}^{o}}$) and ($\cos $${{156}^{o}}$ and $\cos $${{132}^{o}}$) pair of terms together or any other pairs. However, solving any two random pair of terms may not lead to desirable results. In this case, we group ($\cos $ ${{12}^{o}}$ and $\cos $${{132}^{o}}$) and ($\cos $${{84}^{o}}$ and $\cos $${{156}^{o}}$) together. The reason behind this is that after solving, we would get the angles in familiar terms. This would be more clear as solve this question below-

= ($\cos $ ${{12}^{o}}$ and $\cos $${{132}^{o}}$)+ ($\cos $${{84}^{o}}$ and $\cos $${{156}^{o}}$)
= 2$\cos $$\dfrac{{{12}^{o}}+{{132}^{o}}}{2}$$\cos $$\dfrac{{{12}^{o}}-{{132}^{o}}}{2}$ + 2$\cos $$\dfrac{{{84}^{o}}+{{156}^{o}}}{2}$$\cos $$\dfrac{{{84}^{o}}-{{156}^{o}}}{2}$
= 2 $\cos $${{72}^{o}}$$\cos $$(-{{60}^{o}})$+2 $\cos $${{120}^{o}}$$\cos $$(-{{36}^{o}})$
Now, $\cos $ (-x) = $\cos $(x), Thus, we have,
= 2 $\cos $${{72}^{o}}$$\cos $${{60}^{o}}$+2 $\cos $${{120}^{o}}$$\cos $${{36}^{o}}$ -- (A)
Further, $\sin $(90-x) =$\cos $ (x)
Thus, $\cos $${{72}^{o}}$=$\sin $${{(90-72)}^{o}}$= $\sin $${{18}^{o}}$
Substituting this value in (A), we get,
= 2$\sin $${{18}^{o}}$$\cos $${{60}^{o}}$+2$\cos $${{120}^{o}}$$\cos $${{36}^{o}}$

Thus, we were able to get familiar terms by clubbing (cos ${{12}^{o}}$ and cos${{132}^{o}}$) and (cos${{84}^{o}}$ and cos${{156}^{o}}$) together since, we know the numeric values of all these sine and cosine values.
Now using the values of $\sin $${{18}^{o}}$, $\cos $${{36}^{o}}$, $\cos $${{60}^{o}}$and $\cos $${{120}^{o}}$
=$\left( 2\times \dfrac{\sqrt{5}-1}{4}\times \dfrac{1}{2} \right)+\left( 2\times \dfrac{-1}{2}\times \dfrac{\sqrt{5}+1}{4} \right)$
=$\left( \dfrac{\sqrt{5}-1}{4} \right)-\left( \dfrac{\sqrt{5}+1}{4} \right)$
=$-\dfrac{1}{2}$
According to the question,
$\cos {{12}^{o}}+\cos {{84}^{o}}+\cos {{156}^{o}}+\cos {{132}^{o}}=-\dfrac{1}{a}$
Thus,
$-\dfrac{1}{2}$=$-\dfrac{1}{a}$
Thus, a = 2.
Hence, the final answer is a = 2.

Note: While solving trigonometric expressions, it is always important to know the numeric values of sine and cosine of following angles- ${{0}^{o}},{{18}^{o}},{{30}^{o}},{{36}^{o}},{{45}^{o}},{{60}^{o}},{{90}^{o}}$. While solving, we can also group different terms to solve the question. Although the final answer would be the same, it would be more difficult to arrive at the final answer since we would have to manipulate the terms more to get the same answer.