We already have used the tan(θ+θ)=tanθ+tanθ all over 1-tanθtanθ
equation.. What can i do? Are their answers the same? why or why not?
Update:It's this:
It is given that sin2θ=sin(θ+θ)
then using the addition of sines,
sinθcosθ+cosθsinθ then the final is
2sinθcosθ
Another is cos2θ=cos(θ+θ)
cosine addition: cosθcosθ-sinθsinθ
then cos²θ-sin²θ then using a Pythagorean identity, cos²θ-(1-cos²θ) and it will be equal to 2cos²θ-1
another for cos2θ is by terms of sine
cosine addition: cosθcosθ-sinθsinθ
then cos²θ-sin²θ then using a Pythagorean identity, (1-sin²θ)-sin²θ and this will be equal to 1-2sin²θ
i think there's another one but i forgot..
then, using tanθ=tan(θ+θ)
=(tanθ+tanθ)/1-tanθtanθ
=2tanθ/1-tan²θ
what will happen to tan2θ when the ratio of sin and cos will be used? Will the answer of (tanθ+tanθ)/1-tanθtanθ be the same with sin2θ/cos2θ? How? (Using the givens above i.e. sin2θ, cos2θ) My classmates told me that the value is repeating and repeating. Did they do something wrong?
Answers & Comments
Verified answer
1. sinθ = -4/5 then cosθ = -3/5 ... tanθ = 4/3
sin2θ = 2sinθcosθ = 2(-4/5)(-3/5)
cos2θ = 1 - 2sin^2θ = 1 - 2(-4/5)^2
tan2θ = 2tanθ / (1 - tan^2θ) = 2(4/3) / (1 - 16/9)
simplify ...
you can do the next one...
note that 8^2 + 15^2 = 17^2
The answers will definitely be the same.
Here's a short proof for your notes,
tan2θ = tan(θ+θ) = (tanθ + tanθ) / ( 1 - tanθtanθ) = 2tanθ/(1-tan²θ)
sin2θ/cos2θ = 2sinθcosθ/(cos²θ - sin²θ)
Divide numerator and denominator by cos²θ,
sin2θ/cos2θ = [2sinθcosθ/cos²θ] / [ (cos²θ - sin²θ)/cos²θ ] = 2tanθ / (1-tan²θ)
You're asking why tan(2θ) = (tanθ + tanθ)/(1 - tanθtanθ) ?
I'm going to switch θ into x for convenience. Start with what you know and see what you can manipulate further, it is after all just algebra.
tan(2x) = sin(2x)/cos(2x) = 2sin(x)cos(x)/(cos^2(x) - sin^2(x))
Now divide top and bottom by cos^2(x)
tan(2x) = 2sin(x)/cos(x) /(1 - tan^2(x)) = 2tan(x)/(1 - tan(x)tan(x)) = (tan(x) + tan(x))/(1 - tan(x)tan(x))
Yes.
Edit: Hi Angela: I believe you already got this problem down packed, and on your own I must say. I went over every single formula here and they all checked OK. I think if you just relax and let it sink in for a day, you will be just fine.
1)Will the answer of (tanθ+tanθ)/1-tanθtanθ be the same with sin2θ/cos2θ? yes:
(tanθ+tanθ)/(1-tanθtanθ) = sin2θ/cos2θ
(tanθ+tanθ)/(1-tanθtanθ) = 2tanθ/(1 - tan²θ)
= 2tanθ/((1 - tanθ)(1 + tanθ))
= tan2θ
= sin2θ/cos2θ
sin2θ=sin(θ+θ)â sin(a+b) = sinacosb+cosasinb ==> a=b= θ
=sinθcosθ+cosθsinθ â factor out sinθcosθ
=sinθcosθ(1 + 1)
=2sinθcosθ
cos2θ =cos(θ+θ)âcos(a+b)=cosacosb-cosbcosa==>a=b=θ
= cosθcosθ-sinθsinθ
= cos²θ-sin²θ
=cos²θ-(1-cos²θ)
=cos²θ - 1 + cos²θ
=2cos²θ -1
sin2θ/cos2θ ==>
=2sinθcosθ / (2cos²θ -1)
= 2sinθcosθ/(1 - 2sin²θ)
=2tanθ/(1 - tan²θ)
=tan2θ
Regards.
You should learn to do it yourself, this looks like homework. What will you do in the exam? Take a phone browser to put the exam questions on Yahoo Answers?
tan2θ=sin2θ/cos2θ is an identity. It is true for all θ for which each side is defined.
By definition tan2Ã = sin2Ã/cos2Ã
If you want alternative formulae you can expand sin2Ã and cos2Ã and divide them. Ultimately this will reduce to the other formula you already have (and is how you can prove it).
You can calculate sin2Ã and cos2Ã and divide them and you should get the same answer as using the formula. Is that what you're asking?
tan2θ=sin2θ/cos2θ by definition
answer is in the question only by definition we get the same ans