trigonometry
posted by Anonymous .
find cos(θ)ˏsin(θ)ˏtan(θ), if cot (2θ)=5/12 with 0≤2θ≤π  

If cot(2Ø) = 5/12 , then
tan(2Ø) = 12/5 and 2Ø is in either I or III
tan 2Ø = 2tanØ/(1  tan^2 Ø)
let tanØ = u
then 12/5 = 2u/(1  u^2) , (for easier typing)
10u = 12  12u^2
12u^2 + 10u  12 = 0
6u^2 + 5u  6 = 0
(2u + 3)(3u  2) = 0
u = 3/2 or u = 2/3
tanØ = 3/2 or tan u = 2/3
case 1:
tanØ = 3/2> Ø in II or IV
make a sketch of a right rightangled triangle
hypotenuse = √(3^2+2^2) = √13
IN II: sinØ = 2/√13 , cosØ = 3/√13, tanØ = 3/2
IN IV: sinØ = 2/√13 , cosØ = 3/√13, tanØ = 3/2
case 2:
tanØ = 2/3> Ø in I or III
make a new sketch, hypotenuse is still √13
IN I: sinØ = 3/√13, cosØ = 2/√13 , tanØ = 2/3
IN III: sinØ = 3/√13 , cosØ = 2/√13, tanØ = 2/3
check my arithmetic, should have written it out on paper first.