Question #defee

1 Answer
Mar 15, 2016

sin 2t = - 6/13

Explanation:

P is in the Quadrant II.
We have: angle t = (pi/2 + angle a).
Angle a is defined by tan a = 2/3 --> sin a = 2/sqrt(4 + 9) = 2/sqrt13 -->
cos^2 a = 1 - sin^2 a = 1 - 4/13 = 9/13 --> cos a = 3/sqrt13.
Find the trig functions of t.
sin t = sin (a + pi/2) = cos a = 3/sqrt13
cos t = cos (pi/2 + a) = - sin a = - 2/sqrt13
sin 2t = 2sin t.cos t = - (2/sqrt13)(3/sqrt13) = - 6/13
cos^2 2t = 1 - sin^2 t = 1 - 36/169 = 133/169
cos 2t = - sqrt133/13. (2t is in Quadrant III)
sin (t - 150) = sin t.cos 150 - sin 150.cos t.
= - (3/sqrt13)(-sqrt3/2) - (1/2)(-2/sqrt13) =
= - ((3sqrt3)/(2sqrt13)) + (2/(2sqrt13)) = (2 - 3sqrt3)/(2sqrt13)
cos (225 - t) = cos 225.cos t + sin 225.sin t =
cos 225 = -cos 45 = -sqrt2/2 ; and sin 225 = -sin 45 = -sqrt2/2