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Question: How do you find the six Trig values of \[ - {450^ \circ }?...

How do you find the six Trig values of [ - {450^ \circ }?

Explanation

Solution

We will put −450∘ - {450^ \circ }directly as the value of theta inside the six trigonometric expressions and we will try to find the value of these values.
The six trigonometric expressions are Sine, Cosine, Tangent, Cotangent, Secant and Cosecant.
And, these are expressed in the following short form:
sin⁡θ\sin \theta , cos⁡θ\cos \theta , tan⁡θ\tan \theta , cot⁡θ\cot \theta , sec⁡θ\sec \theta , and cos⁡ecθ\cos ec\theta .
Now, the rule of all, sin, cos and tan defines the following points:
1.1.The value of all trigs has positive value in the first quadrant.
2.$$$$\sin \theta ,\cos ec\theta are only positive in the second quadrant.
3.$$$$\tan \theta ,\cot \theta are only positive in the third quadrant.
4.$$$$\cos \theta ,\sec \theta are only positive in the fourth quadrant.
Now, we also know that sin⁡(−θ)=−sin⁡θ\sin ( - \theta ) = - \sin \theta and, tan⁡(−θ)=−tan⁡θ\tan ( - \theta ) = - \tan \theta .
We also know that:
cos⁡ecθ=1sin⁡θ\cos ec\theta = \dfrac{1}{{\sin \theta }}, cot⁡θ=1tan⁡θ\cot \theta = \dfrac{1}{{\tan \theta }}, sec⁡θ=1cos⁡θ\sec \theta = \dfrac{1}{{\cos \theta }} and, tan⁡θ=sin⁡θcos⁡θ\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}.

Complete step by step answer:
First of all we will put θ=450∘\theta = {450^ \circ } in the above trigonometric values.
So, we can write it as:
sin⁡(−450∘),cos⁡(−450∘),tan⁡(−450∘),cot⁡(−450∘),sec⁡(−450∘),cos⁡ec(−450∘)\sin ( - {450^ \circ }),\cos ( - {450^ \circ }),\tan ( - {450^ \circ }),\cot ( - {450^ \circ }),\sec ( - {450^ \circ }),\cos ec( - {450^ \circ }).
Now, we will calculate the sin⁡(−450∘)\sin ( - {450^ \circ }) first.
So, we can write it as following:
sin⁡(−450∘)=−sin⁡(450∘)\sin ( - {450^ \circ }) = - \sin ({450^ \circ }).
So, if we look at it closely, then we can say that we have taken a full clockwise movement in the quadrant of 360∘{360^ \circ } and then we move another 90∘{90^ \circ }from there.
Which says that we will reach at the end of the first quadrant .
But the first quadrant value of sin⁡(450∘)\sin ({450^ \circ }) will be the same as sin⁡(90∘)\sin ({90^ \circ }), which is equal to 11.
So, sin⁡(−450∘)=−1\sin ( - {450^ \circ }) = - 1.
Similarly, we will calculate the value of cos⁡(−450∘)\cos ( - {450^ \circ }).
So, in cos⁡(−450∘)\cos ( - {450^ \circ }) if we move450∘{450^ \circ } in the anti-clockwise direction, again we will be left at the end of the first quadrant.
So, the value of cos⁡(−450∘)\cos ( - {450^ \circ }) is also equal to the value of cos⁡(−90∘)\cos ( - {90^ \circ }), which is equal to 00.
So, we can find the value of tan⁡(−450∘)\tan ( - {450^ \circ }) using the formula tan⁡θ=sin⁡θcos⁡θ\tan \theta = \dfrac{{\sin \theta }}{{\cos \theta }}.
So, tan⁡(−450∘)=sin⁡(−450∘)cos⁡(−450∘)=−10=∞.\tan ( - {450^ \circ }) = \dfrac{{\sin ( - {{450}^ \circ })}}{{\cos ( - {{450}^ \circ })}} = \dfrac{{ - 1}}{0} = \infty .
Again, we can find the value of cot⁡(−450∘)\cot ( - {450^ \circ }) using the formula cot⁡θ=1tan⁡θ\cot \theta = \dfrac{1}{{\tan \theta }}.
So, cot⁡(−450∘)=1tan⁡(−450∘)=1∞=0.\cot ( - {450^ \circ }) = \dfrac{1}{{\tan ( - {{450}^ \circ })}} = \dfrac{1}{\infty } = 0.
Similarly,
sec⁡(−450∘)=1cos⁡(−450∘)=10=∞.\sec ( - {450^ \circ }) = \dfrac{1}{{\cos ( - {{450}^ \circ })}} = \dfrac{1}{0} = \infty .
And, cos⁡ec(−450∘)=1sin⁡(−450∘)=1−1=−1.\cos ec( - {450^ \circ }) = \dfrac{1}{{\sin ( - {{450}^ \circ })}} = \dfrac{1}{{ - 1}} = - 1.

Therefore, the six trigonometric (Sin, Cos, Tan, Cot, Cosec and, Sec) of −450∘ - {450^ \circ }are −1,0,∞,0,−1,∞. - 1,0,\infty ,0, - 1,\infty .

Note: Points to remember: If the given value of theta is equal to the one-fourth of the value of the total measurement of the quadrant, then we can easily use the quadrants to calculate the values of the trigs values.