Radians and Sine help?

Question

Okay, we didn't have time in class today to get to this section but have to do the homework. The question is...

In each case (I'm gonna give 2) write the following angles between 0 and (pi/2) inclusive or between 0 degrees or 90 degrees, as appropriate. For example, sin (5pi/4) = -sin (pi/4).

I do not understand what this is actually wanting me to do. Two of the problems are...

a) sin (3pi/2) Would that be -sin (pi/2) ???


b) sin 300 degrees. I have no idea


If you know any websites, that may be better than just giving the answer. Thanks

Answer 1

Question a)
sin (3π/2) = - 1 = - sin(π/2)
(as you suggest)

Question b)
sin 300° = - sin 60° = - √3 / 2

Remember
sin | ALL
---------
tan | cos

Answer 2

What happens is this: sine is positive in quadrants 1 and 2, and negative in q3 and q4. Since 3 pi / 2 is straight down and its sine is -1, you are correct that it's - sin pi/2 since sin pi/2 = 1.

As for 300 degrees, subtract it from 360 to get the reference angle ( angle between 0 and 90) getting 60, then note that since 300 is in q4, sin 300 is negative, so answer is - sin 60

Answer 3

This is the first site listed of many which I obtained by typing 'trig ratios and quadrants' in the Yahoo Quick Search bar:
http://library.thinkquest.org/20991/alg2/trig.html#ratios

It explains how the 'reference angle' for angles outside the range 0-90deg. is found, and how to find the ratios for these angles.

(a) You are right.
(b) sin(300deg) = -sin(60deg).

Answer 4

sin (3pi/2) = sin(pi+pi/2)
= sin(pi)cos(pi/2) + cos(pi)sin(pi/2)
= 0 -1
= -1

sin(300) = sin (360 - 60)
= sin(360)cos(60) - cos(360)sin(60)
= 0 - sin(60)
= - sqrt(3)/2 = - 1.732/2
= - 0.866