Question

How do you find the $6$ trigonometric functions for $600$degrees?

Answer 1

In this question we have to find the $6$ trigonometric ratios at ${{600}^{\circ }}$ which implies we have to find $\sin \left( {{600}^{\circ }} \right),\cos \left( {{600}^{\circ }} \right),\tan \left( {{600}^{\circ }} \right),\cot \left( {{600}^{\circ }} \right),\sec \left( {{600}^{\circ }} \right)$ and $\csc \left( {{600}^{\circ }} \right)$. We will first find out what is the coterminal angle to ${{600}^{\circ }}$and then find the value of $\sin \left( {{600}^{\circ }} \right)$ and $\cos \left( {{600}^{\circ }} \right)$ , and then use these two functions to find the remaining functions.

Complete step-by-step solution:
We will first see how many loops we can get from the angle ${{600}^{\circ }}$.
Each loop is completed after ${{360}^{\circ }}$ therefore, angle can be divided and written as:
$\Rightarrow {{360}^{\circ }}+{{240}^{\circ }}$
Now the angle ${{240}^{\circ }}$can be written as a difference of ${{360}^{\circ }}$ as:
$\Rightarrow {{360}^{\circ }}+\left( {{360}^{\circ }}-{{120}^{\circ }} \right)$
On taking the term out as common and simplifying, we get:
$\Rightarrow 2\times {{360}^{\circ }}-{{120}^{\circ }}$
Now since two loops are completed fully, the coterminal angle is:
$\Rightarrow -{{120}^{\circ }}$
Now we will find the value of $\sin \left( -{{120}^{\circ }} \right)$ and $\cos \left( -{{120}^{\circ }} \right)$.
We know that $\sin \left( \theta \right)=-\sin \theta$ and $\cos \left( -\theta \right)=\cos \theta$ therefore, we can say that:
$\Rightarrow \sin \left( {{600}^{\circ }} \right)=-\sin \left( {{120}^{\circ }} \right)$
$\Rightarrow \cos \left( {{600}^{\circ }} \right)=\cos \left( {{120}^{\circ }} \right)$
Now we the angle in both the functions can be written as:
$\Rightarrow -\sin \left( {{120}^{\circ }} \right)=-\sin ({{60}^{\circ }}+{{60}^{\circ }})$
$\Rightarrow \cos \left( {{120}^{\circ }} \right)=\cos ({{60}^{\circ }}+{{60}^{\circ }})$
Now we know the double angle formula which is $\sin (a+b)=\sin a\cos b+\cos a\sin b$ and $\cos (a+b)=\cos a\cos b-\sin a\sin b$ therefore, on using the formula, we get:
$\Rightarrow -\sin \left( {{120}^{\circ }} \right)=-\left( \sin \left( {{60}^{\circ }} \right)\cos \left( {{60}^{\circ }} \right)+\sin \left( {{60}^{\circ }} \right)\cos \left( {{60}^{\circ }} \right) \right)$
$\Rightarrow \cos \left( {{120}^{\circ }} \right)=\cos \left( {{60}^{\circ }} \right)\cos \left( {{60}^{\circ }} \right)-\sin \left( {{60}^{\circ }} \right)\sin \left( {{60}^{\circ }} \right)$
Now, we know that $\sin \left( {{60}^{\circ }} \right)=\dfrac{\sqrt{3}}{2}$ and $\cos \left( {{60}^{\circ }} \right)=\dfrac{1}{2}$, on substituting the values, we get:
$\Rightarrow -\sin \left( {{120}^{\circ }} \right)=-\left( \dfrac{\sqrt{3}}{2}\times \dfrac{1}{2}+\dfrac{\sqrt{3}}{2}\times \dfrac{1}{2} \right)$
$\Rightarrow \cos \left( {{120}^{\circ }} \right)= \left( \dfrac{1}{2}\times \dfrac{1}{2}-\dfrac{\sqrt{3}}{2}\times \dfrac{\sqrt{3}}{2} \right)$
On simplifying, we get:
$\Rightarrow -\sin \left( {{120}^{\circ }} \right)=-\dfrac{\sqrt{3}}{2}$
$\Rightarrow \cos \left( {{120}^{\circ }} \right)=-\dfrac{1}{2}$
Now from these two functions, we can evaluate the rest of the functions.
We know $\sec \theta =\dfrac{1}{\cos \theta }$ therefore, $\sec \left( {{600}^{\circ }} \right)=\dfrac{1}{\cos \left( {{600}^{\circ }} \right)}$.
On substituting the value, we get:
$\Rightarrow \sec \left( {{600}^{\circ }} \right)=\dfrac{1}{\left( -1/2 \right)}=-2$
We know $\csc \theta =\dfrac{1}{\sin \theta }$ therefore, $\csc \left( {{600}^{\circ }} \right)=\dfrac{1}{\sin \left( {{600}^{\circ }} \right)}$.
On substituting the value, we get:
$\Rightarrow \csc \left( {{600}^{\circ }} \right)=\dfrac{1}{\left( -\sqrt{3}/2 \right)}=-\dfrac{2\sqrt{3}}{3}$
We know $\tan \theta =\dfrac{\sin \theta }{\cos \theta }$ therefore, $\tan \left( {{600}^{\circ }} \right)=\dfrac{\sin \left( {{600}^{\circ }} \right)}{\cos \left( {{600}^{\circ }} \right)}$.
On substituting the value, we get:
$\Rightarrow \tan \left( {{600}^{\circ }} \right)=\dfrac{\left( -\sqrt{3}/2 \right)}{\left( 1/2 \right)}=-\sqrt{3}$
And we know $\cot \theta =\dfrac{\cos \theta }{\sin \theta }$ therefore, $\cot \left( {{600}^{\circ }} \right)=\dfrac{\cos \left( {{600}^{\circ }} \right)}{\sin \left( {{600}^{\circ }} \right)}$.
On substituting the value, we get:
$\Rightarrow \cot \left( {{600}^{\circ }} \right)=\dfrac{\left( 1/2 \right)}{\left( -\sqrt{3}/2 \right)}= -\dfrac{\sqrt{3}}{3}$
Which are the required solutions.

Note: It is to be remembered that to add two or more fractions, the denominator of both of them should be the same, if the denominator is not the same, the lowest common multiple known as L.C.M should be taken.
The various trigonometric identities and formulae should be remembered while doing these types of sums. the various Pythagorean identities should also be remembered while doing these type of questions
To simplify any given equation, it is good practice to convert all the identities into $\sin$ and $\cos$ for simplifying.