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December 18, 2014

December 18, 2014

Posted by **David** on Thursday, November 28, 2013 at 3:26pm.

- Trigonometry -
**Reiny**, Thursday, November 28, 2013 at 5:20pmif 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 -
**Steve**, Thursday, November 28, 2013 at 8:39pmbut, of course, cos2θ will not be -3/√13, since cot2θ>0 and 2θ<π

- Steve - Trigonometry -
**Reiny**, Thursday, November 28, 2013 at 9:31pmSteve, 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

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