Asked by Val
Determine the value of tan2θ when sinθ=12/13 and pi/2 < θ < pi
Answers
Answered by
mathhelper
Several ways to do this, this way came to mind
Did you recognize the 5-12-13 right-angled triangle?
So in quad II if sinθ = 12/13 , then cosθ = -5/13
tan 2θ = sin 2θ/cos 2θ
= 2sinθcosθ/(cos^2 θ - sin^2 θ)
= 2(12/13)(-5/13) / (25/169 - 144/169)
= (-120/169) / (-119/169)
= -120/-119
= 120/119
Did you recognize the 5-12-13 right-angled triangle?
So in quad II if sinθ = 12/13 , then cosθ = -5/13
tan 2θ = sin 2θ/cos 2θ
= 2sinθcosθ/(cos^2 θ - sin^2 θ)
= 2(12/13)(-5/13) / (25/169 - 144/169)
= (-120/169) / (-119/169)
= -120/-119
= 120/119
Answered by
oobleck
or, since you know that tanθ = -12/5 then
tan2θ = 2tanθ / (1 - tan^2θ)
= 2(-12/5) / (1 - (12/5)^2)
= 120/119
tan2θ = 2tanθ / (1 - tan^2θ)
= 2(-12/5) / (1 - (12/5)^2)
= 120/119
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.