Question

Question #d9cff

Answer 1

`+- (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)`