# Trigonometry

posted by on .

if cot 2θ = 5/12 with 0≤2θ≤π find cos θ, sin θ, tan θ

• Trigonometry - ,

if cot 2Ø = 5/12
then tan 2Ø = 12/5

recognizing the familar 5-12-13 right-angled triangle, we would say

sin 2Ø/cos 2Ø = 12/5 = (12/13) / (5/13)

so sin 2Ø = 12/13 ----> 2sinØcosØ = 12/13
and cos 2Ø = 2cos^ Ø -1 = 5/13
2cos^2 Ø = 18/13
cos^2 Ø = 9/13
cosØ = ± 3/√13

if cosØ = 3/√13
in 2sinØcoØ = 12/13
sinØ(3/√13) = 6/13
sinØ = (6/√13)(√13/3) = 2√13/13 or 2/√13
tanØ = 2/3

if cosØ = -3/√13
(only the signs will change, same method)
sinØ = -2/√13
tanØ = 2/3

• Trigonometry - ,

but, of course, cos2θ will not be -3/√13, since cot2θ>0 and 2θ<π

• Steve - Trigonometry - ,

Steve, you are right, should have looked at that original restriction more carefully.
( must have automatically assumed the usual
0 ≤ Ø ≤ 2π )

So you will find my other solution comical, where I assumed that domain and got solutions in all 4 quadrants :

...

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 right-angled 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 = 2/√13, cosØ = 3/√13 , tanØ = 2/3
IN III: sinØ = -2/√13 , cosØ = -3/√13, tanØ = 2/3