Question

How do you find the exact values ${{\cos }^{-1}}\left( -\dfrac{\sqrt{2}}{2} \right)$?

Answer 1

We need to simplify the value in the bracket. Find the values of the cos by using the right angle triangle formula. According to the right angle triangle formula, the cosine is equal to the adjacent side divided by hypotenuse. We must take the values in both the positive axis and the negative axis.
${{\cos }^{-1}}\left( -\dfrac{\sqrt{2}}{2} \right)$ is the given term. We need to find the exact values for the given term.

Complete step-by-step solution:
Now let’s take the value in the bracket.
$\Rightarrow \left( -\dfrac{\sqrt{2}}{2} \right)$
Now we need to simplify the value
$\Rightarrow \left( -\dfrac{1}{\sqrt{2}} \right)$
Now we simplified ${{\cos }^{-1}}\left( -\dfrac{\sqrt{2}}{2} \right)$ as $\cos \left( \dfrac{1}{\sqrt{2}} \right)$
Here,
If $\cos \left( \dfrac{1}{\sqrt{2}} \right)$ is equal to $\theta$ then
$\Rightarrow \cos \theta =\left( -\dfrac{1}{\sqrt{2}} \right)$
If the values of $\theta \in \left[ 0,2\pi \right)$. Then we will have the possibilities in two quadrants. One quadrant with the negative x-axis and positive y-axis also. The other quadrant with the negative x-axis and negative y-axis.
In the quadrant of the positive y-axis and negative x-axis for the angle of ${{45}^{\circ }}$.
The hypotenuse value will be $\sqrt{2}$, the value of the opposite side will be $1$and the value of the adjacent side will be $-1$.
In the quadrant of the negative x-axis and negative y-axis for the angle of ${{45}^{\circ }}$.
The hypotenuse value will be $\sqrt{2}$, the value of the opposite side will be $-1$and the value of the adjacent side will be $-1$. Here the hypotenuse value is the same as the value of the positive axis because the value is in the square root.
Now let’s consider the values of the positive quadrant.
${{180}^{\circ }}+{{45}^{\circ }}={{225}^{\circ }}$ Is one value and the other value will be ${{180}^{\circ }}-{{45}^{\circ }}={{135}^{\circ }}$
Now let's the above values in the radians notations.
Then we get
$\pi -\dfrac{\pi }{4}=\dfrac{3\pi }{4}$, $\pi +\dfrac{\pi }{4}=\dfrac{5\pi }{4}$
Therefore the values of ${{\cos }^{-1}}\left( -\dfrac{\sqrt{2}}{2} \right)$ are ${{135}^{\circ }},{{225}^{\circ }}$

Note: We calculated the values by using the quadrant. There are four quadrants. In the first quadrant, all the trigonometric values will be positive. In the second quadrant sin and cosec values will be positive. In the third quadrant, tan and cot values will be positive. In the fourth quadrant cos and sec, values will be positive.