Expert: Steve Holleran - 3/21/2007

Question
I need to know how to do a problem, and I don't even know where to start :(

sin(theta)=9/10, theta is in quadrant 2
sin(theta + pi/6)=

i have no idea :(

Answer
Hi Brad,

I'm going to use A instead of theta here, because its easier to write.

Okay, to get started, diagram an angle in standard position (vertex at the origin, initial side along the positive x-axis) which has its terminal side in the second quadrant.  You should have an acute reference angle, call it B, between the terminal side of your angle and the negative x-axis.  Draw a triangle (the reference triangle) by dropping a perpendicular from the terminal side of the angle to the negative x-axis.

Now, since sin A = 9/10, and sin = opposite/hypotenuse,
label the side across from the reference angle B as 9 and the hypotenuse (the terminal side of your original angle) as 10. Using pythagorean theorem, the side of the triangle along the negative x-axis becomes sqrt(100 - 81) = sqrt(19).  You have to label it with a - sign also, since we're in quadrant II:
                      y
                 |\ 10|
              9  | \  |
                 |  \ |
                 | B \|
           -------------------------------X
               -sqrt19|

Now, for sin(A + pi/6), you need to use the formula for the sin of (x+y): sin x cos y + cos x sin y.  So here we have:

sin(A + pi/6) = sin A cos pi/6 + cos A sin pi/6

             =  9/10 * sqrt(3)/2 + -sqrt(19)/10 * 1/2

             =  9*sqrt(3) / 20 - sqrt(19) / 20

or,              [9 sqrt(3) - sqrt(19)] / 20

which is approximately .5615

I hope I didn't confuse you with the explanation or the diagram, and I hope this helps you out.

Steve Holleran