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September 18, 2014

September 18, 2014

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

- trigonometry -
**Reiny**, Thursday, November 28, 2013 at 9:26amIf cot(2Ų) = 5/12 , then

tan(2Ų) = 12/5 and 2Ų is in either I or III

tan 2Ų = 2tanŲ/(1 - tan^2 Ų)

let tanŲ = 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

tanŲ = -3/2 or tan u = 2/3

case 1:

tanŲ = -3/2--> Ų in II or IV

make a sketch of a right right-angled triangle

hypotenuse = √(3^2+2^2) = √13

**IN II: sinŲ = 2/√13 , cosŲ = -3/√13, tanŲ = -3/2**

IN IV: sinŲ = -2/√13 , cosŲ = 3/√13, tanŲ = -3/2

case 2:

tanŲ = 2/3---> Ų in I or III

make a new sketch, hypotenuse is still √13**IN I: sinŲ = 3/√13, cosŲ = 2/√13 , tanŲ = 2/3**

IN III: sinŲ = -3/√13 , cosŲ = -2/√13, tanŲ = 2/3

check my arithmetic, should have written it out on paper first.

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