Question

The value of $cos({12^ \circ }) + cos({84^ \circ }) + cos({156^ \circ }) + cos({132^ \circ })$ is ___________.

Answer 1

Hint : : Here we are only given the trigonometric function cosine, denoted as $cos()$. When we wish to calculate the sum of multiple trigonometric functions we must find a way to group them. We will do rough calculations in our mind to find out the best groupings in such a way that we are able to use direct trigonometric values to solve.
Formulas used:
$\cos x + \cos y = 2\cos (\dfrac{{x + y}}{2})\cos (\dfrac{{x - y}}{2})$
$\cos ( - x) = \cos x$
$\cos (x) = \sin (90 - x)$
$\cos (90 + x) = - \sin (x)$
$\cos ({36^ \circ }) = \dfrac{{(\sqrt 5 + 1)}}{4}$
$\sin ({18^ \circ }) = \dfrac{{(\sqrt 5 - 1)}}{4}$

Complete step-by-step answer :
From the question we see that all the functions that are being operated upon are cosines. This decreases our confusion in choosing the right formula, because when multiple cosine functions are added, we generally use the formula:
$\cos x + \cos y = 2\cos (\dfrac{{x + y}}{2})\cos (\dfrac{{x - y}}{2})$, to find the answer.
Clearly this formula can only be used for two cosine functions at a time, but in the question we are given four different cosine functions so we need to group them in pairs (in two) to be able to use this formula.
Now grouping must be done carefully, because it determines the ease to solve the problem. If we group between some terms, it can be difficult to reach the final answer.
From the question $cos({12^ \circ }) + cos({84^ \circ }) + cos({156^ \circ }) + cos({132^ \circ })$ take a look at:
$cos({12^ \circ }),cos({132^ \circ })$ and $cos({156^ \circ }),cos({84^ \circ })$
These pairs look like the best option to be clubbed together.
Now moving on to solving the given operation, given question is:
$\Rightarrow cos({12^ \circ }) + cos({84^ \circ }) + cos({156^ \circ }) + cos({132^ \circ })$
Grouping in pairs;
$= [cos({12^ \circ }) + cos({132^ \circ })] + [cos({156^ \circ }) + cos({84^ \circ })]$
Applying the formula of $(\cos x + \cos y)$ for each pair;
$= \left[ {2 \times \cos \dfrac{{{{(12 + 132)}^ \circ }}}{2}\cos \dfrac{{{{(12 - 132)}^ \circ }}}{2}} \right] + \left[ {2 \times \cos \dfrac{{{{(156 + 84)}^ \circ }}}{2}\cos \dfrac{{{{(156 - 84)}^ \circ }}}{2}} \right]$
Now simplifying the fractions, we get;
$= \left[ {2 \times \cos ({{72}^ \circ })\cos ( - {{60}^ \circ })} \right] + \left[ {2 \times \cos ({{120}^ \circ })\cos ({{36}^ \circ })} \right]$
Here we see that one of the angles: $\cos ( - 60)$, but by the formula $\cos ( - x) = \cos x$, we can rewrite the previous simplification as;
$= \left[ {2 \times \cos ({{72}^ \circ })\cos ({{60}^ \circ })} \right] + \left[ {2 \times \cos ({{120}^ \circ })\cos ({{36}^ \circ })} \right]\; \to (i)$
We do not know the direct value of $\cos (72{}^ \circ )$, so let us change it in terms of sine function:
$\cos (72{}^ \circ ) = \sin ({(90 - 72)^ \circ })$
$\Rightarrow \cos (72{}^ \circ ) = \sin ({18^ \circ })$
Also we do not have a direct value for $\cos ({120^ \circ })$, so it will be:
$\cos ({120^ \circ }) = \cos ({(90 + 30)^ \circ })$
$\Rightarrow \cos ({120^ \circ }) = - \sin ({30^ \circ })$
Now we move on to substitute $\sin ({18^ \circ }), - \sin ({30^ \circ })$ in place of $\cos (72{}^ \circ )$ and $\cos (120{}^ \circ )$ respectively in equation $(i)$;
$= \left[ {2 \times \sin ({{18}^ \circ })\cos ({{60}^ \circ })} \right] + \left[ {2 \times - \sin ({{30}^ \circ })\cos ({{36}^ \circ })} \right]$
Then putting the values
$\sin ({18^ \circ }) = \dfrac{{(\sqrt 5 - 1)}}{4}$, $\cos ({36^ \circ }) = \dfrac{{(\sqrt 5 + 1)}}{4}$, $\cos ({60^ \circ }) = \dfrac{1}{2}$ and $- \sin ({30^ \circ }) = - \dfrac{1}{2}$ into the equation;
$\Rightarrow cos({12^ \circ }) + cos({84^ \circ }) + cos({156^ \circ }) + cos({132^ \circ }) = \left[ {2 \times \dfrac{{(\sqrt 5 - 1)}}{4} \times \dfrac{1}{2}} \right] + \left[ {2 \times - \dfrac{1}{2} \times \dfrac{{(\sqrt 5 + 1)}}{4}} \right]$
Simplifying gives us;
$\Rightarrow cos({12^ \circ }) + cos({84^ \circ }) + cos({156^ \circ }) + cos({132^ \circ }) = \left[ {\dfrac{{(\sqrt 5 - 1)}}{4}} \right] + \left[ { - \dfrac{{(\sqrt 5 + 1)}}{4}} \right]$

Therefore the final value for the given question is:
$cos({12^ \circ }) + cos({84^ \circ }) + cos({156^ \circ }) + cos({132^ \circ }) = - \dfrac{1}{2}$
So, the correct answer is $- \dfrac{1}{2}$”.

Note : Grouping terms while solving operations between trigonometric functions can be tricky. When we group functions, we need to group them in a way we can apply the direct and familiar trigonometric values to arrive at the final answer. Once we practice multiple questions of the similar kind, we will know which groups can give the easiest solution.