How do you find the six trigonometric functions of 420 degrees?

Hint:
We start solving the problem by considering the sine of the given angle. We then make use of the facts that $ \sin \left( {{360}^{\circ }}+\theta \right)=\sin \theta $ and $ \sin {{60}^{\circ }}=\dfrac{\sqrt{3}}{2} $ to get the value of $ \sin {{420}^{\circ }} $ . We then consider the cosine of the given angle. We then make use of the facts that $ \cos \left( {{360}^{\circ }}+\theta \right)=\cos \theta $ and $ \cos {{60}^{\circ }}=\dfrac{1}{2} $ to get the value of $ \cos {{420}^{\circ }} $ . We then make use of the facts $ \tan \theta =\dfrac{\sin \theta }{\cos \theta } $ , $ \operatorname{cosec}\theta =\dfrac{1}{\sin \theta } $ , $ \sec \theta =\dfrac{1}{\cos \theta } $ and $ \cot \theta =\dfrac{1}{\tan \theta } $ to get the value of the other trigonometric functions at the given angle.

Complete step by step answer:
According to the problem, we are asked to find the values of the six trigonometric functions of 420 degrees.
Let us first find the value of the sine function of 420 degrees.
So, we have $ \sin \left( {{420}^{\circ }} \right)=\sin \left( {{360}^{\circ }}+{{60}^{\circ }} \right) $ ---(1).
We know that $ \sin \left( {{360}^{\circ }}+\theta \right)=\sin \theta $ . Let us use this result in equation (1).
 $ \Rightarrow \sin \left( {{420}^{\circ }} \right)=\sin {{60}^{\circ }} $ ---(2).
We know that $ \sin {{60}^{\circ }}=\dfrac{\sqrt{3}}{2} $ . Let us use this result in equation (2).
 $ \Rightarrow \sin \left( {{420}^{\circ }} \right)=\dfrac{\sqrt{3}}{2} $ ---(3).
Now, let us first find the value of the cosine function of 420 degrees.
So, we have $ \cos \left( {{420}^{\circ }} \right)=\cos \left( {{360}^{\circ }}+{{60}^{\circ }} \right) $ ---(4).
We know that $ \cos \left( {{360}^{\circ }}+\theta \right)=\cos \theta $ . Let us use this result in equation (4).
 $ \Rightarrow \cos \left( {{420}^{\circ }} \right)=\cos {{60}^{\circ }} $ ---(5).
We know that $ \cos {{60}^{\circ }}=\dfrac{1}{2} $ . Let us use this result in equation (5).
 $ \Rightarrow \cos \left( {{420}^{\circ }} \right)=\dfrac{1}{2} $ ---(6).
Now, we know that $ \tan \theta =\dfrac{\sin \theta }{\cos \theta } $ .
From equations (3) and (6) we get $ \tan \left( {{420}^{\circ }} \right)=\dfrac{\sin \left( {{420}^{\circ }} \right)}{\cos \left( {{420}^{\circ }} \right)}=\dfrac{\dfrac{\sqrt{3}}{2}}{\dfrac{1}{2}} $ .
 $ \Rightarrow \tan \left( {{420}^{\circ }} \right)=\sqrt{3} $ ---(7).
We know that $ \operatorname{cosec}\theta =\dfrac{1}{\sin \theta } $ .
From equation (3), we get $ \operatorname{cosec}\left( {{420}^{\circ }} \right)=\dfrac{1}{\sin \left( {{420}^{\circ }} \right)}=\dfrac{1}{\dfrac{\sqrt{3}}{2}} $ .
 $ \Rightarrow \operatorname{cosec}\left( {{420}^{\circ }} \right)=\dfrac{2}{\sqrt{3}} $ .
We know that $ \sec \theta =\dfrac{1}{\cos \theta } $ .
From equation (6), we get $ \sec \left( {{420}^{\circ }} \right)=\dfrac{1}{\cos \left( {{420}^{\circ }} \right)}=\dfrac{1}{\dfrac{1}{2}} $ .
 $ \Rightarrow \sec \left( {{420}^{\circ }} \right)=2 $ .
We know that $ \cot \theta =\dfrac{1}{\tan \theta } $ .
From equation (3), we get $ \cot \left( {{420}^{\circ }} \right)=\dfrac{1}{\tan \left( {{420}^{\circ }} \right)}=\dfrac{1}{\sqrt{3}} $ .
 $ \Rightarrow \cot \left( {{420}^{\circ }} \right)=\dfrac{1}{\sqrt{3}} $ .

Note:
 Whenever we get this type of problems, we try to make use of the trigonometric identities and then try to write the angles into the ones between $ {{0}^{\circ }} $ and $ {{90}^{\circ }} $ to make the calculation simpler. We should not make calculation mistakes while solving this problem. We can also find the equivalent value of the given angle by subtracting it with the multiple of $ {{360}^{\circ }} $ to get them between $ {{0}^{\circ }} $ and $ {{90}^{\circ }} $ . Similarly, we can expect problems to find the values of all trigonometric functions at –300 degrees.