Call x the arctan (2/3)arctan(23) --> tan x = 2/3tanx=23
cos^2 x = 1/(1 + tan^2 x) = 1/( 1 + 4/9) = 9/13cos2x=11+tan2x=11+49=913
cos x = +- 3/sqrt13cosx=±3√13
sin x = 1 - cos^2 x = 1 - 9/13 = 4/13sinx=1−cos2x=1−913=413
sin x = +- 2/sqrt13sinx=±2√13
Since tan x = 2/3tanx=23 --> 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) =sin(5π6−x)=sin(5π6).cosx−sinx.cos(5π6)=
= (1/2)(3/sqrt13) - (2/sqrt13)(-sqrt3/2) = (3 + 2sqrt3)/(2sqrt13)=(12)(3√13)−(2√13)(−√32)=3+2√32√13
b. If sin x and cos x are both negative:
sin ((5pi)/6 - x) = (1/2)(-3/sqrt13) - (-2/sqrt13)(- sqrt3/2) = sin(5π6−x)=(12)(−3√13)−(−2√13)(−√32)=
= - (3 + 2sqrt3)/(2sqrt13)=−3+2√32√13
