since π < θ < 3π/2 (QIII),
sinθ = -5/13
tanθ = 5/12
Now just plug those into your double-angle formulae.
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
4 answers
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
sin2θ = 2sinθcosθ = 2(-5/13)(-12/13) = 120/169
cos2θ = 2cos^2θ-1 = 2(144/169)-1 = 119/169
tan2θ = sin2θ/cos2θ = 120/119
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
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
I messed up on my sign (QII)
So, Reiny, how did you get 5/12? That was -tanØ
So, Reiny, how did you get 5/12? That was -tanØ
1 answer