Expert: Sherman D. - 12/8/2007

Question
Hi, I'm taking a brief course in trigonometric and i'm having a hard time finding the value for trigonometric functions. For instance, in one problem it said that the 2 values for sin(theta)=-(square root 3)/2 is 4pi/3 and 5pi/3. But in some problems one of the values is the reference angle (in this case it would be pi/3.) My question is how do you know when to use the reference angle and also how to find the values if the reference angle isn't one of the values? Thank you.

Answer
Here's how

sin(A) = +n if 0° < A < pi , Quadrants I and II (x,y) or (-x,y)

sin(A) = -n if 0° > A > pi , Quadrants III and IV, (x,-y) or (-x,-y)

cos(A) = +n if pi/2 > A > 3pi/2 , Quadrants I and IV, (x,y) or (x,-y)

cos(A) = -n if pi/2 < A < 3pi/2 , Quadrants II and III, (-x,y) or (-x,-y)

tan(A) = n if 0° < A < pi/2 or pi < A < 3pi/2, Quadrants I and III, (x,y) or (-x,-y)

tan(A) = -n if 0° > A > (3pi/2) or pi > A > pi/2, Quadrants II and IV, (-x,y) or (x,-y)

Depending if it should be -n or +n, just subtract/add the value you get win you say A = sin^-1(-sqrt(3)/2) from pi or 0.

sin(A) = -sqrt(3)/2
A = sin^-1(-sqrt(3)/2)
A = -pi/3 , since this is suppose to be negative, you just take that and put 2pi - pi/3 and also pi + pi/3

so

2pi - (pi/3) = (6pi - pi)/3 = (5pi/3)
pi + (pi/3) = (3pi + pi)/3 = (4pi/3)

So the 2 answers as you can see are (5pi/3) and (4pi/3)

The reason i put the Quadrants and the (x,y) stuff, is because of this

sin(A) = y/r
cos(A) = x/r
tan(A) = y/x

to find out cos(A), tan(A), or sin(A), you need to use
x^2 + y^2 = r^2

More info found at http://id.mind.net/~zona/mmts/trigonometryRealms/TrigFuncPointDef/TrigFuncPointD...