Question #d9cff

1 Answer
Nghi N.
Mar 6, 2017

+- (3 + 2sqrt3)/(2sqrt13)

Explanation:

Call x the arctan (2/3) --> tan x = 2/3
cos^2 x = 1/(1 + tan^2 x) = 1/( 1 + 4/9) = 9/13
cos x = +- 3/sqrt13
sin x = 1 - cos^2 x = 1 - 9/13 = 4/13
sin x = +- 2/sqrt13
Since tan x = 2/3 --> x could be in Quadrant 1 or Quadrant 3.
If x is in Q. 1, sin x and cos x are both positive
If x is in Q.3, sin x and cos x are both negative.
Use trig identity sin ( a - b) = sin a.cos b - sin b.cos a
a. If sin x and cos x are both positive
sin ((5pi)/6 - x) = sin ((5pi)/6).cos x - sin x.cos ((5pi)/6) =
= (1/2)(3/sqrt13) - (2/sqrt13)(-sqrt3/2) = (3 + 2sqrt3)/(2sqrt13)
b. If sin x and cos x are both negative:
sin ((5pi)/6 - x) = (1/2)(-3/sqrt13) - (-2/sqrt13)(- sqrt3/2) =
= - (3 + 2sqrt3)/(2sqrt13)

Related questions
See all questions in Trigonometric Functions of Any Angle
Impact of this question
1071 views around the world
You can reuse this answer
Creative Commons License