Question

How do you find the six trigonometric values of $–450$ degrees?

Answer 1

Hint: We will first write $– 450$ in terms of $– 360$ and some $x$, then we will write all the trigonometric functions with it, then we will use the general formulas of them.

Complete step by step answer:
We are given that we are required to find the sex trigonometric values of – 450 degrees.
Now, we will use the fact that:-
In the fourth quadrant, only cosine and secant are positive.
Therefore, cos (-x) = cos x and sec (-x) = sec x but sin (-x) = - sin x, cosec (-x) = - cosec x, tan (-x) = -tan x and cot (-x) = - cot x.
Using this, we will get: $\sin ( - {450^ \circ }) = - \sin ({450^ \circ }),\cos ( - {450^ \circ }) = \cos ({450^ \circ }),\tan ( - {450^ \circ }) = - \tan ({450^ \circ })$
$\cos ec( - {450^ \circ }) = - \cos ec({450^ \circ }),\sec ( - {450^ \circ }) = \sec ({450^ \circ }),\cot ( - {450^ \circ }) = - \cot ({450^ \circ })$
We can definitely write – 450 as – (360 + 90).
Therefore, $\sin ( - {450^ \circ }) = - \sin ({360^ \circ } + {90^ \circ })$ ……………….(1)
First, we will do the solution for the sine of – 450 degrees.
Now, we will use the fact that $\sin ({360^ \circ } + \theta ) = \sin \theta $
Putting the theta as 90 degrees, we will then obtain the following equation:-
$ \Rightarrow \sin ({360^ \circ } + {90^ \circ }) = \sin {90^ \circ }$
Now, we already know that $\sin {90^ \circ } = 1$. Therefore, we will obtain the following equation:-
$ \Rightarrow \sin ({360^ \circ } + {90^ \circ }) = 1$
Putting this above equation in equation number 1, we will then obtain the following equation:-
Therefore, $\sin ( - {450^ \circ }) = - 1$
Now, we already know that cosecant is inverse of sine, therefore, $\cos ec( - {450^ \circ }) = \dfrac{1}{{\sin ( - {{450}^ \circ })}} = - 1$
Now, we will do the solution for the cos of – 450 degrees.
Since we established that $\cos ( - {450^ \circ }) = \cos ({450^ \circ })$.
Now, we will use the fact that $\cos ({360^ \circ } + \theta ) = \cos \theta $
Therefore, $\cos ( - {450^ \circ }) = \cos ({360^ \circ } + {90^ \circ })$ ……………….(2)
Putting the theta as 90 degrees, we will then obtain the following equation:-
$ \Rightarrow \cos ({360^ \circ } + {90^ \circ }) = \cos {90^ \circ }$
Now, we already know that $\cos {90^ \circ } = 0$. Therefore, we will obtain the following equation:-
$ \Rightarrow \cos ({360^ \circ } + {90^ \circ }) = 0$
Putting this above equation in equation number 2, we will then obtain the following equation:-
Therefore, $\cos ( - {450^ \circ }) = 0$
Now, we already know that secant is inverse of cosine, therefore, $\sec ( - {450^ \circ }) = \dfrac{1}{{\cos ( - {{450}^ \circ })}}$.
Therefore, the secant of – 450 degrees is not defined.
Now, we will do the solution for the tan of – 450 degrees.
Since we established that $\tan ( - {450^ \circ }) = - \tan ({450^ \circ })$.
Now, we will use the fact that $\tan ({360^ \circ } + \theta ) = \tan \theta $
Therefore, $\tan ( - {450^ \circ }) = - \tan ({360^ \circ } + {90^ \circ })$ ……………….(3)
Putting the theta as 90 degrees, we will then obtain the following equation:-
$ \Rightarrow \tan ({360^ \circ } + {90^ \circ }) = \tan {90^ \circ }$
Now, we already know that $\tan {90^ \circ }$ is not defined. Therefore, we will obtain the following equation:-
$ \Rightarrow \tan ({360^ \circ } + {90^ \circ })$ is not defined.
Putting this above equation in equation number 3, we will then obtain the following equation:-
Therefore, $\tan ( - {450^ \circ })$ is not defined.
Now, we already know that cotangent is inverse of tangent, therefore, $\cot ( - {450^ \circ }) = \dfrac{1}{{\cot ( - {{450}^ \circ })}} = \dfrac{1}{\infty } = 0$.

Note: The students must commit to memory that we can judge whether a trigonometric function is positive or negative in any quadrant using the “ADD SUGAR TO COFFEE” where A stands for all and it implies that in first quadrant, all the trigonometric ratios are positive, S stands for sine and it implies that in second quadrant, sine is positive, T stands for tangent and it implies that in third quadrant, tangent is positive and C stands for cosine and it implies that in fourth quadrant, cosine is positive.
The students must commit to memory the facts that:-
$\sin ( - {450^ \circ }) = - \sin ({450^ \circ }),\cos ( - {450^ \circ }) = \cos ({450^ \circ }),\tan ( - {450^ \circ }) = - \tan ({450^ \circ })$
$\cos ec( - {450^ \circ }) = - \cos ec({450^ \circ }),\sec ( - {450^ \circ }) = \sec ({450^ \circ }),\cot ( - {450^ \circ }) = - \cot ({450^ \circ })$