Cos (-120)

[casper22]

---

Members don't see this ad.

This is from MATH Destroyer 2010 Test 6 - Q7

Q - Which value is equal to cos (-120)

A. sin 60
B. cos 60
C. -sin 60
D. cos 30
E. -cos60

Destroyer says the answer is E and the reasoning for that is -120 is in 3rd quadrant..which I completely agree and also the fact that in that quadrant its reference angle is 60 and cos is negative in 3rd Quadrant...I agree with all these facts...my question is why cant the answer be sin 60...since sin is positive in 3rd Quadrant and the reference still remains 60...

What am I missing...

 

[invictusx]

This is from MATH Destroyer 2010 Test 6 - Q7

Q - Which value is equal to cos (-120)

A. sin 60
B. cos 60
C. -sin 60
D. cos 30
E. -cos60

Destroyer says the answer is E and the reasoning for that is -120 is in 3rd quadrant..which I completely agree and also the fact that in that quadrant its reference angle is 60 and cos is negative in 3rd Quadrant...I agree with all these facts...my question is why cant the answer be sin 60...since sin is positive in 3rd Quadrant and the reference still remains 60...

What am I missing...

sin is negative in the 3rd quadrant; positive in the 1st and 2nd

 

[Jab1113]

sin is negative in the 3rd quadrant; positive in the 1st and 2nd

Q1 - A
Q2 - S
Q3 - T
Q4 - C

"All Students Take Calculus"

 

[casper22]

SWEET!!...Thanks guys...I got my quadrants messed up...

 

[UFiA0389]

but isnt the reference angle: the angle minus 180 for the 3rd quadrant? why isnt the reference angle 300? bcuz -120-180 = 300. Or do we ignore the neg sign when subtracting 180 from the angle?

 

[mh0000]

but isnt the reference angle: the angle minus 180 for the 3rd quadrant? why isnt the reference angle 300? bcuz -120-180 = 300. Or do we ignore the neg sign when subtracting 180 from the angle?

your reference angles will always be acute. So the magnitude of the ref. angle will be 180-120=60

 

[UFiA0389]

your reference angles will always be acute. So the magnitude of the ref. angle will be 180-120=60

Oh so thats for every quadrant? Just subtract the angle from 180?

 

[mh0000]

Oh so thats for every quadrant? Just subtract the angle from 180?

Just find the acute angle from the x-axis depending on which quadrant you are in

 

[UFiA0389]

Thanks!