Thursday

December 8, 2016
Posted by **Anonymous** on Wednesday, November 27, 2013 at 11:42pm.

- trigonometry -
**Reiny**, Thursday, November 28, 2013 at 9:26amIf 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 = 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.