could someone please help explain how find the exact values would appreciate it. Ex: cos 135 degrees
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Hello, are you familiar with the unit circle?
For example, \[\cos (30)=\frac{\sqrt{3}}{2}\]
hey and uhm no not that much but I know what it is. I know the values and etc like you labeled cos 30
Okay, what you really want to do is checkout the unit circle, at each major point on the circle is an x,y coordinate
sin(x) is equivalent to the y value and cos(x) is equivalent to the x value
okay so suppose the angle here I labeled as an example is 135 but what if its like above 360 for example 460 degrees?
the easiest way is to convert to radians and subtract 2 pi
Do you know how to do that?
Would mean alot if you could help just began this and already confused and got a test tomorrow.
I know how to convert to radians not the rest.
on the unit circle, take a look at 135 deg
cos(x) is equivalent of the x value, you see \[-\frac{\sqrt{2}}{2}\]
that is the exact value
yes x is is that in the pair of coordinates
Yeah
That's correct
could you please do me a favor
Sure
I did this on my own my study guide please verify or criticize on my wrong doings, I was to once again find the exact value for an angle of 125 degrees so I made a graph and did 180-125 and got 55 and since tan 1255 is in the second quadrant and negative i got the final answer of -tan 55 degrees. would that be correct?
tan 125* not 1255 sorry
Yes, that is
phew! thanks. but one more question could you please help me with an example if the angle was greater than 360 degrees?
Whats your question?
cos 850 degrees how would I solve that for the exact value?
main point or question being what to do if the value is greater than 360
Round to the nearest tenth. Well, that is if your dong trigonometry. Are you dong Trigonometry?
yes I am.
Alright, well anything that is equal to 360 degrees is a circle. But, cos85 = -36.8292. What you do next is that you round -36.8292 to the nearest 10th.
uhm I wrote 850 but we still round?
You have to solve cos 850 to get your answer..
alright I think I got the idea thanks :)
No problem.
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