anonymous
  • anonymous
cos(2tan-1x) rewrite as an algebraic expression
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
there is something wrong in the expression!!
anonymous
  • anonymous
the tan-1 is inverse tan not minus 1
anonymous
  • anonymous
oh I see

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
sorry about that
anonymous
  • anonymous
Well, arctan(x) is just an angle in a right angled triangle, then tan of which is x. So if you draw a triangle where tan(angle) = x, what are the side lengths?
anonymous
  • anonymous
Maybe it makes more sense like this: let α = arctan(x) <=> tan(α) = x Therefore the opposite side over the ajacent side = .... ,
anonymous
  • anonymous
OK I'll do one more step but no further without some sign you are thinking about it yourself: In a RA triangle with one angle α, the opposite side over the adjacent side = x, so let the sides be x and 1, respectively. Therefore, by Pythagoras, the hypotenuse = ...., and so cos α = ....
anonymous
  • anonymous
sorry i was grabbing lunch
anonymous
  • anonymous
OK then. Do you see where I'm going, though?
anonymous
  • anonymous
yeah i understand. a^2 +b^2 = c ^2. what i don't understand is where the 2 comes int0 play
anonymous
  • anonymous
before the tangent
anonymous
  • anonymous
You can treat tan^-1(x) as a normal angle, so if you know what cos(tan^-1(x)) is just use the formula for cos(2a)
anonymous
  • anonymous
im sorry but that doesn't make any sense to me. my brain is peicing together what you're saying
anonymous
  • anonymous
piecing*
anonymous
  • anonymous
1 Attachment
anonymous
  • anonymous
Does that image help explain it better?
anonymous
  • anonymous
hold on one second
anonymous
  • anonymous
You could similarly use cos^2(x) + sin^2(x) = 1 -> 1 + tan^2(a) - 1/cos^2(a) -> cos^2(a) = 1/(1+tan^2(a)) And we know tan(a) = x as a = tan^-1(x)
anonymous
  • anonymous
Let y=arctanx. The original equation then becomes cos2y. Using an identity this becomes cosy^2-siny^2. So your goal here is to obtain cosy^2 and siny^2. So we let y=arctanx. This means tany=x. We square and add 1 to both sides because of the identity tany^2+1=secy^2. So tany^2+1=x^2+1 and using the identity just mentioned, secy^2=x^2+1. Inverse both sides to get cosy^2= 1/ x^2+1. Now to obtain siny, simply multiple cosy (which you get by sqrting the cosy^2 from above) on both sides by tany. And since in the beginning we said tany=x. We get siny=x/sqrt(x^2+1). Subbing everything back into the orignal we get 1-x^2/(x^2+1)
anonymous
  • anonymous
so if i just take what i find for tan (opposite/adjacent) and distribute a 2 before putting it back into terms of cosine?
anonymous
  • anonymous
No, you take what is cos (adj/hyp = 1/sqrt(x^2+1) ) And then use cos(2a) = 2cos^2(a) - 1 = 2(1/sqrt(x^2+1) )^2 -1 = 2/(x^2+1) -1
anonymous
  • anonymous
You know tan(a) = x by definition
anonymous
  • anonymous
okay so cos(2a) got it. so you're just using tan to find the value's of the triangle?
anonymous
  • anonymous
Yes, from which you can work out cos (or sin, but you don't need both here)
anonymous
  • anonymous
ohhhhh okay. tan-1x = a ? so 2tan-1x = 2a right then cos(2a) is what i want?
anonymous
  • anonymous
Yes, indeed :D
anonymous
  • anonymous
ahh thanks let me work it out and ill reply with an answer
anonymous
  • anonymous
cos(2sqrt(1+x^2)/x) ?
anonymous
  • anonymous
Sorry there is a slight error. You use the triangle (or the equations) to work out cos(a) and THEN you use this value of cos(a) in the formula cos(2a) = 2cos^2(a) - 1. You do not need to work out cos or anything complicated, ad you already have the value of cos(a).
anonymous
  • anonymous
my book says the answer is (1-x^2)/(1+x^2)
anonymous
  • anonymous
but idk how to get there
anonymous
  • anonymous
You could similarly use cos^2(x) + sin^2(x) = 1 -> 1 + tan^2(a) - 1/cos^2(a) -> cos^2(a) = 1/(1+tan^2(a)) And we know tan(a) = x as a = tan^-1(x) So cos^2(a) = 1/(1+x^2) and so cos(2a) is just this minus 1. Both would give the same value of cos(a) (and cos^2(a) ) , it just depends how you prefer to think about it
anonymous
  • anonymous
OK, I'll help you with the next bit. From the triangle, you know cos(a) = adj/hyp = 1/sqrt(1+x^2) So cos^2(a) = 1/(1+x^2) Cos(2a) = 2cos^2(a) - 1 = 2/(1+x^2) - 1 = (2 - 1 - x^2)/1+x^2 = (1-x^2)/(1+x^2)
anonymous
  • anonymous
okay yes
anonymous
  • anonymous
Sorry the simplification/fraction look messy but hopefully the brackets make it clear
anonymous
  • anonymous
You could also have got sin^2(a) form the triangle (x^2/(1+x^2)) and used cos(2a) = cos^2(a) - sin^2(a) = (1-x^2)/(1+x^2), it's all the same.
anonymous
  • anonymous
those are pythag identities ?
anonymous
  • anonymous
thats what i was missing. i didin't know i was supposed to be using those to get my answer.
anonymous
  • anonymous
cos(2a) = cos^2(a) - sin^2(a) = 2cos^2(a)-1 = 1-2sin^2(a) This is a specfic version of the cos formulae: cos(a+b) = cos(a)cos(b) - sin(a)sin(b) It is not a pythag identity, and is pretty hard to prove
anonymous
  • anonymous
no no i meant...
anonymous
  • anonymous
cos^2 - 1
anonymous
  • anonymous
that's part of cos^2 + sin^2 = 1 right?
anonymous
  • anonymous
you used that to get the answer?
anonymous
  • anonymous
It depends, you either use that, or the triangle to get the first part - I showed both methods - but you still need cos(2a) later.
anonymous
  • anonymous
okay thanks! off to take my test.
anonymous
  • anonymous
Have fun!
anonymous
  • anonymous
lol sure

Looking for something else?

Not the answer you are looking for? Search for more explanations.