Posted by David on Thursday, November 28, 2013 at 3:26pm.
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
but, of course, cos2θ will not be -3/√13, since cot2θ>0 and 2θ<π
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