if cot 2θ = 5/12 with 0≤2θ≤π find cos θ, sin θ, tan θ
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
To find the values of cos θ, sin θ, and tan θ, we'll start by using the given equation cot 2θ = 5/12.
1. First, let's simplify the given equation. The reciprocal identity for cotangent tells us that cot θ = 1/tan θ. Therefore, we can rewrite the equation as tan 2θ = 12/5.
2. Now, we need to use the double-angle formula for tangent. The formula is tan 2θ = (2tan θ)/(1-tan²θ).
3. Substituting tan 2θ = 12/5 into the formula, we get (2tan θ)/(1-tan²θ) = 12/5.
4. Solving this equation algebraically, we can cross-multiply to get 12(1 - tan²θ) = 10tan θ.
5. Expanding the equation, we have 12 - 12tan²θ = 10tan θ.
6. Rearranging the terms, we have 12tan²θ + 10tan θ - 12 = 0.
To solve this quadratic equation, we can use the quadratic formula:
t = (-b ± √(b² - 4ac))/2a
For our equation, let a = 12, b = 10, and c = -12.
Using the quadratic formula, we have:
tan θ = (-10 ± √(10² - 4(12)(-12)))/(2(12))
Simplifying further, we have:
tan θ = (-10 ± √(100 + 576))/24
tan θ = (-10 ± √676)/24
tan θ = (-10 ± 26)/24
Now, we have two possibilities:
1. tan θ = (26 - 10)/24 = 16/24 = 2/3
2. tan θ = (-26 - 10)/24 = -36/24 = -3/2
To determine which of these is correct, we need to consider the given restriction 0≤2θ≤π.
Since 0≤2θ≤π, this means that the angle 2θ lies in the first or second quadrant.
In the first and second quadrants, the tangent function is positive, so we eliminate the solution tan θ = -3/2.
Therefore, we conclude that tan θ = 2/3.
Now that we have the value of tan θ, we can find cos θ and sin θ using the Pythagorean identities:
1. To find cos θ, we can use the formula:
cos θ = 1/√(1 + tan²θ)
Plugging in the value of tan θ = 2/3, we have:
cos θ = 1/√(1 + (2/3)²)
cos θ = 1/√(1 + 4/9)
cos θ = 1/√((9 + 4)/9)
cos θ = 1/√(13/9)
cos θ = √(9/13)
cos θ = 3/√13
To rationalize the denominator, we multiply the numerator and denominator by √13:
cos θ = (3/√13)(√13/√13)
cos θ = 3√13/13
Therefore, cos θ = 3√13/13.
2. To find sin θ, we can use the formula:
sin θ = tan θ * cos θ
Plugging in the values of tan θ = 2/3 and cos θ = 3√13/13, we have:
sin θ = (2/3) * (3√13/13)
sin θ = 2√13/3.
Therefore, sin θ = 2√13/3.
To summarize the results:
cos θ = 3√13/13
sin θ = 2√13/3
tan θ = 2/3