find cos(θ)ˏsin(θ)ˏtan(θ), if cot (2θ)=5/12 with 0≤2θ≤π - -
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 = 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.
To find the values of cos(θ), sin(θ), and tan(θ), we will first find the value of θ using the given information about cot(2θ).
We are given that cot(2θ) = 5/12 and 0 ≤ 2θ ≤ π.
We know that cot(2θ) is the reciprocal of tan(2θ), so we can find tan(2θ) by taking the reciprocal of 5/12.
tan(2θ) = 1 / (5/12) = 12 / 5
Now, we can use the double angle formula for tangent to find tan(θ):
tan(θ) = tan(2θ) / (1 - tan^2(2θ))
= (12/5) / (1 - (12/5)^2)
= (12/5) / (1 - 144/25)
= (12/5) / (25/25 - 144/25)
= (12/5) / (-119/25)
= -12/119
Now that we have tan(θ), we can use that to find cos(θ) and sin(θ) using the trigonometric identities.
cos(θ) = 1 / sqrt(1 + tan^2(θ))
= 1 / sqrt(1 + (-12/119)^2)
= 1 / sqrt(1 + 144/14161)
= 1 / sqrt(14161 + 144) / 14161
= 1 / sqrt(14305) / 14161
= 14161 / sqrt(14305)
sin(θ) = tan(θ) * cos(θ)
= (-12/119) * (14161 / sqrt(14305))
= (-12 * 14161) / (119 * sqrt(14305))
Therefore, cos(θ) = 14161 / sqrt(14305), sin(θ) = (-12 * 14161) / (119 * sqrt(14305)), and tan(θ) = -12/119.