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Mathematics 15 Online
OpenStudy (anonymous):

1. Using your calculator, complete the table below for the function y = cos x. Find the corresponding y- values rounded to the nearest thousandth. x y 0° -30° -60° -90° -120° -150° -180° -210° -240° -270° -300° -330° -360°   

OpenStudy (anonymous):

We know that cos(-x) = cos x.... apply cos(90-x) , cos(90+x) etc to get the values ;)

OpenStudy (anonymous):

@vikrantg4 , I don't know how to do it

OpenStudy (anonymous):

It's difficult to type all the answers here :P

OpenStudy (anonymous):

just tell the first 3 so i can get an idea

OpenStudy (anonymous):

@vikrantg4

OpenStudy (phi):

do you have a calculator?

OpenStudy (anonymous):

no :S

OpenStudy (phi):

you could use google type in the google search window cos(0 deg)=

OpenStudy (anonymous):

ok

OpenStudy (anonymous):

i got 1

OpenStudy (phi):

that is because the cosine of zero degrees is one.

OpenStudy (anonymous):

ok, so the first one is1, how do i find these second one?

OpenStudy (anonymous):

what should i put?

OpenStudy (phi):

think about it and try it.

OpenStudy (anonymous):

-30 is 0.866025404? right?

OpenStudy (phi):

I would say cos(-30)= 0.866... notice this: type in sqrt(3)/2= does that number look familiar?

OpenStudy (anonymous):

yeah, ok, i realized that if i put '' cos(-30)'' or '' cos(30)'' I get the same number so it doesn't matter if it's negative or positive, right?

OpenStudy (phi):

notice if you type cos(-30)= in google you get 0.15425145 that is because it interprets 30 as being in radians. So be sure to say cos(-30 deg)= and you can try cos(-30 deg) and compare it to cos(30 deg)

OpenStudy (phi):

If you had a calculator, you would have to make sure it was in "degree mode" and not "radian mode"

OpenStudy (anonymous):

ok, well thanks a lot

OpenStudy (phi):

also, the question says rounded to the nearest thousandth. that means to 3 decimals. Look at the 4th digit to the right of the decimal point. if 5 or bigger round up. 0.866025404 rounds to 0.866

OpenStudy (anonymous):

yeah i got that, thanks

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