The exact value of cos(pi/4) is 1/square root 2 but when working out cos (-7pi/4) why is the answer square root 2/2 ?
cos is an even function \[\huge\rm Cos(-x)=\cos(x)\] what is cos at 7pi/4 radi ??
I have no idea, can you explain it to me further. I got pi/4
so how did you get pi/4 ?? :=)
r u allowed to use unit circle ?:=)
by going around the unit circle a lot ?? yes i am
alright
|dw:1440591648001:dw| (x,y) solution where c-coordinate represent cos and y-coordinate = sin so what is cos at 7pi/4 ?
and no pi/4 isn't \[\huge\rm \frac{ 1 }{ \sqrt{2} }\]you can't leave the root at the denominator you have to multiply both top and bottom of the fraction by square root 2
how do i find the value of these coordinates if i am not given the unit circle ??
oh! so i just have to make sure there are no square roots at the denominator ??
and i will get the coordinates i am looking for? thanks!!
yes right \[\textrm {no square root at the denominator }\]
without looking at the unit circle 2 ways 1) familiar with the 30-60-90 and 45-45-90 triangle 2nd) memorize
and for example if they ask `what is the exact value of cos(5pi/4) then cos equal to -sqrt{2} over 2 |dw:1440591987978:dw|
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