come on, man. This is 4th grade addition of fractions.
13/18 = 9/18 + 5/18
now, cos(13π/18) = cos(π/2) cos(5π/18) - sin(π/2) sin(5π/18) = -sin(5π/18)
so y = -5π/18
If you want a positive angle, just add 2π to that.
Given that cos 13pi/18=sin y, first express 13pi/18 as a sum of pi/2 and an angle, and then apply a trigonometric identitiy to determine the measure of angle y
3 answers
13 π / 18 = π / 2 + θ
Subtract π / 2 to both sides
13 π / 18 - π / 2 = θ
13 π / 18 - 9 π / 18 = θ
4 π / 18 = θ
2 ∙ 2 π / 2 ∙ 9 = θ
2 π / 9 = θ
θ = 2 π / 9
13 π / 18 = π / 2 + 2 π / 9
Use identity:
cos ( π / 2 + θ ) = - sin ( θ )
cos ( 13 π / 18 ) = cos ( π / 2 + 2 π / 9 ) = - sin ( 2 π / 9 )
Now:
cos ( 13 π / 18 ) = sin ( y )
- sin ( 2 π / 9 ) = sin ( y )
sin ( y ) = - sin ( 2 π / 9 )
The solutions are:
y = - 2 π / 9 + 2 π n
________________
Since:
sin ( π + θ ) = - sin( θ )
sin ( π + 2 π / 9 ) = - sin ( 2 π / 9 )
sin ( 9 π / 9 + 2 π / 9 ) = - sin ( 2 π / 9 )
sin ( 11 π / 9 ) = - sin ( 2 π / 9 )
y = 11 π / 9 + 2 π n
________________
Subtract π / 2 to both sides
13 π / 18 - π / 2 = θ
13 π / 18 - 9 π / 18 = θ
4 π / 18 = θ
2 ∙ 2 π / 2 ∙ 9 = θ
2 π / 9 = θ
θ = 2 π / 9
13 π / 18 = π / 2 + 2 π / 9
Use identity:
cos ( π / 2 + θ ) = - sin ( θ )
cos ( 13 π / 18 ) = cos ( π / 2 + 2 π / 9 ) = - sin ( 2 π / 9 )
Now:
cos ( 13 π / 18 ) = sin ( y )
- sin ( 2 π / 9 ) = sin ( y )
sin ( y ) = - sin ( 2 π / 9 )
The solutions are:
y = - 2 π / 9 + 2 π n
________________
Since:
sin ( π + θ ) = - sin( θ )
sin ( π + 2 π / 9 ) = - sin ( 2 π / 9 )
sin ( 9 π / 9 + 2 π / 9 ) = - sin ( 2 π / 9 )
sin ( 11 π / 9 ) = - sin ( 2 π / 9 )
y = 11 π / 9 + 2 π n
________________
oops. How ever did I get 5/18?
Luckily, it was Bosnian for the save!
Luckily, it was Bosnian for the save!