Patrick B. answered 03/20/20
Math and computer tutor/teacher
1) sin 165 = sin (120 + 45) = sin120 cos45 + cos120 sin 45 = (sqrt(3)/2)(-sqrt(2)/2)+ (-1/2)(sqrt(2)/2)
= -sqrt(6)/4 - sqrt(2)/4
= -1/4 [ sqrt(6) + sqrt(2)]
2) 2 pi - 7/12 pi = pi*(24-7)/12 = (17/12)*pi
partial fraction decomposition
A/4 + B/3 = 17/12
3A + 4B = 17
A=3 B=2
9/12 pi = 3/4 pi and 2/3 pi makes 17/12 pi
cos ( 3/4 pi) cos (2/3 pi) - sin (3/4 pi) sin (2./3 pi) =
cos (135) cos ( 120) - sin (135) sin (120) =
-sqrt(2)/2 * -1/2 - sqrt(2)/2 * sqrt(3)/2 =
sqrt(2)/4 - sqrt(6)/ 4 =
[ sqrt(2) - sqrt(6) ]/4
(3)
cos(u-v) = cos(u)cos(v) + sin(u)sin(v)
cos(u) = -15/17 in quadrant 3 where it is negative
it is right triangle x = -8, y=-15, hypotenuse=17
sin(u) = -8/17, cos(u) = -15/17, tan(u) = 15/8
sin(v) = 4/5 in quadrant 1 where it is positive
it is triangle x=3,y=4,hypotenuse=5
sin(v) = 3/5, cos(v) = 4/5, tan(v) = 3/4
cos(u-v) = cos(u)cos(v) + sin(u)sin(v) = (-15/17)(4/5) + (-8/17)(3/5)
= -60/85 + -24/85
= -84/85
Mark M.
In 2) -7/12 became 17/12. -7/12 = 1/6 - 3/403/21/20