Question

How do you evaluate sine, cosine and tangent of $ - {840^ \circ }$ without using a calculator?

Answer 1

Hint: We will first write the given angle in terms of ${360^ \circ }$, then we will just mention the general formulas of sine, cosine and tangent of $\left( {{{360}^ \circ } + \theta } \right)$ and thus we have the required answer.

Complete step by step solution:
We are given that we are required to evaluate sine, cosine and tangent of $ - {840^ \circ }$ without using a calculator.
Now, we can write ${840^ \circ }$ as $\left( {2 \times {{360}^ \circ } + {{120}^ \circ }} \right)$ ……….(1)
Let us first find the sine of it:
Now, we know that we have the following formula with us: $\sin \left( { - \theta } \right) = - \sin \theta $.
Therefore, we will get: $\sin \left( { - {{840}^ \circ }} \right) = - \sin \left( {{{840}^ \circ }} \right)$.
Now, using the equation number 1 in the above equation, we will then obtain the following equation:-
$ \Rightarrow \sin \left( { - {{840}^ \circ }} \right) = - \sin \left( {2 \times {{360}^ \circ } + {{120}^ \circ }} \right)$ ………….(2)
Now, we also have a formula given by: $\sin \left( {2n\pi + \theta } \right) = \sin \theta $.
Using this, we will then obtain:-
$ \Rightarrow \sin \left( {2 \times {{360}^ \circ } + {{120}^ \circ }} \right) = \sin {120^ \circ }$
Putting this in equation number 2, we will then obtain the following equation:-
$ \Rightarrow \sin \left( { - {{840}^ \circ }} \right) = - \sin {120^ \circ }$ ……………(3)
Now, we also have a formula given by: $\sin \left( {{{90}^ \circ } + \theta } \right) = \cos \theta $.
Replacing $\theta $ by ${30^ \circ }$, we will then obtain the following equation:-
$ \Rightarrow \sin \left( {{{120}^ \circ }} \right) = \cos {30^ \circ } = \dfrac{{\sqrt 3 }}{2}$
Putting this in equation number 3, we will then obtain:-
\[ \Rightarrow \sin \left( { - {{840}^ \circ }} \right) = - \dfrac{{\sqrt 3 }}{2}\]
Now, we know that we have the following formula with us: $\sin \left( { - \theta } \right) = - \sin \theta $.
Therefore, we will get: $\sin \left( { - {{840}^ \circ }} \right) = - \sin \left( {{{840}^ \circ }} \right)$.
Now, using the equation number 1 in the above equation, we will then obtain the following equation:-
$ \Rightarrow \sin \left( { - {{840}^ \circ }} \right) = - \sin \left( {2 \times {{360}^ \circ } + {{120}^ \circ }} \right)$ ………….(2)
Now, we also have a formula given by: $\sin \left( {2n\pi + \theta } \right) = \sin \theta $.
Using this, we will then obtain:-
$ \Rightarrow \sin \left( {2 \times {{360}^ \circ } + {{120}^ \circ }} \right) = \sin {120^ \circ }$
Putting this in equation number 2, we will then obtain the following equation:-
$ \Rightarrow \sin \left( { - {{840}^ \circ }} \right) = - \sin {120^ \circ }$ ……………(3)
Now, we also have a formula given by: $\sin \left( {{{90}^ \circ } + \theta } \right) = \cos \theta $.
Replacing $\theta $ by ${30^ \circ }$, we will then obtain the following equation:-
$ \Rightarrow \sin \left( {{{120}^ \circ }} \right) = \cos {30^ \circ } = \dfrac{{\sqrt 3 }}{2}$
Putting this in equation number 3, we will then obtain:-
\[ \Rightarrow \sin \left( { - {{840}^ \circ }} \right) = - \dfrac{{\sqrt 3 }}{2}\]
Let us now find the cosine of it:
Now, we know that we have the following formula with us: $\cos \left( { - \theta } \right) = \cos \theta $.
Therefore, we will get: $\cos \left( { - {{840}^ \circ }} \right) = \cos \left( {{{840}^ \circ }} \right)$.
Now, using the equation number 1 in the above equation, we will then obtain the following equation:-
$ \Rightarrow \cos \left( { - {{840}^ \circ }} \right) = \cos \left( {2 \times {{360}^ \circ } + {{120}^ \circ }} \right)$ ………….(4)
Now, we also have a formula given by: $\cos \left( {2n\pi + \theta } \right) = \cos \theta $.
Using this, we will then obtain:-
$ \Rightarrow \cos \left( {2 \times {{360}^ \circ } + {{120}^ \circ }} \right) = \cos {120^ \circ }$
Putting this in equation number 4, we will then obtain the following equation:-
$ \Rightarrow \cos \left( { - {{840}^ \circ }} \right) = \cos {120^ \circ }$ ……………(5)
Now, we also have a formula given by: $\cos \left( {{{90}^ \circ } + \theta } \right) = - \sin \theta $.
Replacing $\theta $ by ${30^ \circ }$, we will then obtain the following equation:-
$ \Rightarrow \cos \left( {{{120}^ \circ }} \right) = - \sin {30^ \circ } = - \dfrac{1}{2}$
Putting this in equation number 5, we will then obtain:-
\[ \Rightarrow \cos \left( { - {{840}^ \circ }} \right) = - \dfrac{1}{2}\]
Let us now find the tangent of it:
Now, we know that we have the following formula with us: $\tan \left( { - \theta } \right) = - \tan \theta $.
Therefore, we will get: $\tan \left( { - {{840}^ \circ }} \right) = - \tan \left( {{{840}^ \circ }} \right)$.
Now, using the equation number 1 in the above equation, we will then obtain the following equation:-
$ \Rightarrow \tan \left( { - {{840}^ \circ }} \right) = - \tan \left( {2 \times {{360}^ \circ } + {{120}^ \circ }} \right)$ ………….(6)
Now, we also have a formula given by: $\tan \left( {2n\pi + \theta } \right) = \tan \theta $.
Using this, we will then obtain:-
$ \Rightarrow \tan \left( {2 \times {{360}^ \circ } + {{120}^ \circ }} \right) = \tan {120^ \circ }$
Putting this in equation number 6, we will then obtain the following equation:-
$ \Rightarrow \tan \left( { - {{840}^ \circ }} \right) = - \tan {120^ \circ }$ ……………(7)
Now, we also have a formula given by: $\tan \left( {{{90}^ \circ } + \theta } \right) = - \cot \theta $.
Replacing $\theta $ by ${30^ \circ }$, we will then obtain the following equation:-
$ \Rightarrow \tan \left( {{{120}^ \circ }} \right) = - \cot {30^ \circ } = - \sqrt 3 $
Putting this in equation number 7, we will then obtain:-
\[ \Rightarrow \tan \left( { - {{840}^ \circ }} \right) = \sqrt 3 \]
Thus, we have the required answer.

Note: The students must notice that we have used a fact that $\pi = {180^ \circ }$ and using this we converted the formulas from the angle in radians to angle in degrees.
We actually have n as 2, $\theta $ as ${120^ \circ }$ and since $\pi = {180^ \circ }$. Putting all this in $\sin \left( {2n\pi + \theta } \right) = \sin \theta ,\cos \left( {2n\pi + \theta } \right) = \cos \theta $ and $\tan \left( {2n\pi + \theta } \right) = \tan \theta $, we have the required expressions.
The students must also commit to memory the following formulas:-
1. $\sin \left( {2n\pi + \theta } \right) = \sin \theta $
2. $\cos \left( {2n\pi + \theta } \right) = \cos \theta $
3. $\tan \left( {2n\pi + \theta } \right) = \tan \theta $
4. $\sin \left( {{{90}^ \circ } + \theta } \right) = \cos \theta $
5. $\cos \left( {{{90}^ \circ } + \theta } \right) = - \sin \theta $
6. $\tan \left( {{{90}^ \circ } + \theta } \right) = - \cot \theta $