Question

Given $\theta =\dfrac{3\pi }{4}$ how does one calculate ${{\cos }^{2}}\theta$,$\cos \left( -\theta \right)$ and $\cos 2\left( \theta \right)$?

Answer 1

From the question it had been given that, $\theta =\dfrac{3\pi }{4}$. Now, we have to observe that the given angle is in which quadrant and with measurements it creates a reference angle. According to the occurrence of theta in the quadrants, the sign of the theta will change.

Complete step-by-step solution:
Now considering from the question we have $\theta =\dfrac{3\pi }{4}$ .
As here the value of $\theta$ is greater than $\dfrac{\pi }{2}$ it lies in the ${{2}^{nd}}$ quadrant.
We clearly know that the given angle is in quadrant $2$.
This angle creates a reference triangle with leg $\sqrt{2}$, leg $\sqrt{2}$, hypotenuse $2$.
Since $\theta =\dfrac{3\pi }{4}$ is in the quadrant $2$ $\left( \dfrac{\pi }{2}\le \dfrac{3\pi }{4}\le \pi \right)$, the cosine function is negative.
Therefore, $\cos \left( \theta \right)=-\dfrac{\sqrt{2}}{2}$
Now, we have to do the squaring on both sides of the above equation.
By squaring on both sides of the above equation, we will get the below equation,
${{\cos }^{2}}\left( \theta \right)={{\left( -\dfrac{\sqrt{2}}{2} \right)}^{2}}$
$\Rightarrow {{\cos }^{2}}\left( \theta \right)=\dfrac{1}{2}$
Hence, we got the value of ${{\cos }^{2}}\left( \theta \right)=\dfrac{1}{2}$
Because of $-\theta$ occurs in the quadrant three, $\cos \left( -\theta \right)$ is negative, and since it is a $45$ degrees or $\dfrac{\pi }{4}$ angle, it also has the same reference triangle, so $\cos \left( -\theta \right)=-\dfrac{\sqrt{2}}{2}$ as well
Therefore, we got the value of $\cos \left( -\theta \right)=-\dfrac{\sqrt{2}}{2}$
Now, $2\left( \theta \right)=2\left( \dfrac{3\pi }{4} \right)$
$\Rightarrow 2\left( \theta \right)=\dfrac{3\pi }{2}$
Using the unit circle, we know the coordinates of a point on the circle are $\left( \cos ,\sin \right)$, and $\dfrac{3\pi }{2}$ is located on the y axis,
Therefore, $\cos 2\left( \theta \right)=0$.

Note: While answering questions of this type we should be sure with the calculations and concept. Here we use the concept that 1 complete angle measures ${{360}^{\circ }}$ if it is divided into 4 equal parts each part will have ${{90}^{\circ }}$. Quadrant is defined as the ${{\dfrac{1}{4}}^{th}}$ part of the whole plane. So we can say that the first quadrant ranges from ${{0}^{\circ }}\text{ to }{{90}^{\circ }}$ and the second quadrant ranges from ${{90}^{\circ }}\text{ to 18}{{0}^{\circ }}$and the third quadrant ranges from ${{180}^{\circ }}\text{ to 27}{{0}^{\circ }}$and the fourth quadrant ranges from ${{270}^{\circ }}\text{ to 36}{{0}^{\circ }}$.