Find sin2theta, cos2theta, and tan2theta for costheta = -12/13 in pi < theta < 3pi/2.
My answers:
20111/24336 for tan2theta.
144/169 for cos2theta
-120/338 for sin2theta
So we know Ø is in quad III , where both sine and cosine are negative
given : cosØ = -12/13 and recognizing the 5,12,13 right-angled triangle we know that
sinØ = -5/13
sin 2Ø
= 2sinØcosØ
= 2(-5/13)(-12/13)
= 120/169
cos 2Ø
= cos^2 Ø - sin^2 Ø
= 144/169 - 25/169 = 119/169
tan 2Ø = sin 2Ø/cos 2Ø
= (120/169) / (119/169)
= 120/119
= 5/12
How and from where did you get those crazy numbers in your answers
since π < θ < 3π/2 (QIII),
sinθ = -5/13
tanθ = 5/12
Now just plug those into your double-angle formulae.
Not sure I like your values:
sin2θ = 2sinθcosθ = 2(-5/13)(-12/13) = 120/169
cos2θ = 2cos^2θ-1 = 2(144/169)-1 = 119/169
tan2θ = sin2θ/cos2θ = 120/119
I messed up on my sign (QII)
So, Reiny, how did you get 5/12? That was -tanØ
To find sin(2θ), cos(2θ), and tan(2θ) given cos(θ) = -12/13 for θ in the range π < θ < 3π/2, we need to use the double angle formulas.
1. Finding sin(2θ):
The formula for sin(2θ) is sin(2θ) = 2sin(θ)cos(θ).
Since cos(θ) = -12/13, we need to first find sin(θ). To do so, we can use the Pythagorean identity sin²(θ) + cos²(θ) = 1. Solving for sin(θ), we have:
sin(θ) = √(1 - cos²(θ))
sin(θ) = √(1 - (-12/13)²)
sin(θ) = √(1 - 144/169)
sin(θ) = √(169/169 - 144/169)
sin(θ) = √(25/169)
sin(θ) = 5/13 (taking the positive value since θ is in the specified range)
Now, we can substitute sin(θ) and cos(θ) into the formula for sin(2θ):
sin(2θ) = 2(5/13)(-12/13)
sin(2θ) = -120/169
So, sin(2θ) = -120/169
2. Finding cos(2θ):
The formula for cos(2θ) is cos(2θ) = cos²(θ) - sin²(θ).
From our previous calculations, we know that sin(θ) = 5/13 and cos(θ) = -12/13. Substituting these values into the formula:
cos(2θ) = (-12/13)² - (5/13)²
cos(2θ) = 144/169 - 25/169
cos(2θ) = 119/169
So, cos(2θ) = 119/169
3. Finding tan(2θ):
The formula for tan(2θ) is tan(2θ) = (2tan(θ))/(1 - tan²(θ)).
Since we already know sin(θ) = 5/13 and cos(θ) = -12/13, we can calculate tan(θ) using the formula tan(θ) = sin(θ)/cos(θ):
tan(θ) = (5/13)/(-12/13)
tan(θ) = -5/12
Now, we can substitute tan(θ) into the formula for tan(2θ):
tan(2θ) = (2(-5/12))/(1 - (-5/12)²)
tan(2θ) = (2(-5/12))/(1 - 25/144)
tan(2θ) = (-10/12)/(1 - 25/144)
tan(2θ) = (-10/12)/(119/144)
tan(2θ) = (-10/12) * (144/119)
tan(2θ) = - 120/169
So, tan(2θ) = -120/169
Therefore, the answers are:
sin(2θ) = -120/169
cos(2θ) = 119/169
tan(2θ) = -120/169