Here's the question you clicked on:
znimon
Evaluate the expression without using a calculator.
\[\sin (\pi/3) + \cos (\pi/3)\]
An entry level explanation would be useful.
Pi/3=60º as is well known cos60º=1/2 and sin60º=sqrt3/2
Should I have converted them to degrees then?
allow me to disagree with @myko this question has nothing to do with degrees. forget degrees it has to do with numbers find the sine and cosine from the coordinates on the unit circle
if it is more clear for you, then yes, if not just stay with radians
do not start a trig problem by converting numbers to degrees, it is a bad habit and will mess you up later if you are working with numbers, stick with numbers
degrees are just, sometimes, more evident to place a apoint in the circle. But I agree with @satellite73 that it is more strait with rad
sine and cosine are functions of numbers, not angles. they correspond to the functions of angles if the angles are measured in radians
alright that seems to make sense. Will you walk me through what you mean by "find the sine and cosine from the coordinates on the unit circle?"
look at the last page of the cheat sheet i sent locate the point on the unit circle corresponding to \(\frac{\pi}{3}\) 2x ^{5}+x ^{3}-7x+14
corresponding to \(\frac{\pi}{3}\)
alright I see an ordered pair that looks suspiciously like sin and cos that you mentioned earlier
you should see the ordered pair \((\frac{1}{2},\frac{\sqrt{3}}{2})\) the first coordinate is \(\cos(\frac{\pi}{3})\) and the second coordinate is \(\sin(\frac{\pi}{3})\)
So what if this is on a test and I'm not allowed the cheat sheet?
This one seems fairly easy now I know this forms a 60 30 90 triangle but what about something more irregular or do these always form 30-60-90 or 45-45-90 triangles?
(The ones they ask me to solve without a calculator.)