Question

How do you express $\cos \left( \dfrac{27\pi }{8} \right)$as a trig function of an angle in Quadrant $1$?

Answer 1

In this question we have the trigonometric function of $\cos x$ in which the angle $x$ has the value of $\left( \dfrac{27\pi }{8} \right)$ which we have to simplify while considering the solution to be in the first quadrant. We will first convert the angle into a sum of $2\pi +\alpha$ , and since we know that $\cos \left( 2\pi +\alpha \right)=\cos \left( \alpha \right)$, we will simplify the expression. Then we will convert the angle into the sum of $\pi +\beta$, and since we know that in the first quadrant $\cos \left( \pi +\beta \right)=-\cos \left( \beta \right)$, we will simplify it and get the required solution.

Complete step-by-step solution:
We have the trigonometric function given to us as:
$\Rightarrow \cos \left( \dfrac{27\pi }{8} \right)$
Now we have to simplify the angle. We know that $27\pi /8>2\pi$ therefore, we will try to express it as a sum of $2\pi +\alpha$.
We can see that $\dfrac{27\pi }{8}=\dfrac{11\pi }{8}+2\pi$ therefore, on substituting, we get:
$\Rightarrow \cos \left( \dfrac{11\pi }{8}+2\pi \right)$
Now we know that $\cos \left( 2\pi +\alpha \right)=\cos \left( \alpha \right)$ therefore, on using the result, we get:
$\Rightarrow \cos \left( \dfrac{11\pi }{8} \right)$
We know that $11\pi /8>2\pi$ therefore, we will try to express it as a sum of $\pi +\beta$.
We can see that $\dfrac{11\pi }{8}=\pi +\dfrac{3\pi }{8}$ therefore, on substituting, we get:
$\Rightarrow \cos \left( \pi +\dfrac{3\pi }{8} \right)$
Now in the question it has been mentioned that we have to find the angle in the first quadrant. We know that in the first quadrant $\cos \left( \pi +\beta \right)=-\cos \left( \beta \right)$ therefore, on using this result, we get:
$\Rightarrow -\cos \left( \dfrac{3\pi }{8} \right)$, which is the required solution.

Note: Basic trigonometric formulas should be remembered to solve these types of sums. It is to be remembered which trigonometric functions are positive and negative in what quadrants. The formula used over here is for $\cos x$, the other formulas for the $\sin x$ and $\tan x$ should be remembered.
When you add $\pi$ to any angle, its position on the graph reverses, and whenever you add $2\pi$ to any angle, it reaches the same point after a complete rotation.