cos(2tan-1x) rewrite as an algebraic expression

- anonymous

cos(2tan-1x) rewrite as an algebraic expression

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- schrodinger

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- anonymous

there is something wrong in the expression!!

- anonymous

the tan-1 is inverse tan not minus 1

- anonymous

oh I see

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- anonymous

sorry about that

- 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

Maybe it makes more sense like this:
let α = arctan(x) <=> tan(α) = x
Therefore the opposite side over the ajacent side = .... ,

- 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

sorry i was grabbing lunch

- anonymous

OK then. Do you see where I'm going, though?

- anonymous

yeah i understand. a^2 +b^2 = c ^2. what i don't understand is where the 2 comes int0 play

- anonymous

before the tangent

- 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

im sorry but that doesn't make any sense to me. my brain is peicing together what you're saying

- anonymous

piecing*

- anonymous

##### 1 Attachment

- anonymous

Does that image help explain it better?

- anonymous

hold on one second

- 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

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

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

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

You know tan(a) = x by definition

- anonymous

okay so cos(2a) got it. so you're just using tan to find the value's of the triangle?

- anonymous

Yes, from which you can work out cos (or sin, but you don't need both here)

- anonymous

ohhhhh okay. tan-1x = a ? so 2tan-1x = 2a right then cos(2a) is what i want?

- anonymous

Yes, indeed :D

- anonymous

ahh thanks let me work it out and ill reply with an answer

- anonymous

cos(2sqrt(1+x^2)/x) ?

- 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

my book says the answer is (1-x^2)/(1+x^2)

- anonymous

but idk how to get there

- 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

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

okay yes

- anonymous

Sorry the simplification/fraction look messy but hopefully the brackets make it clear

- 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

those are pythag identities ?

- anonymous

thats what i was missing. i didin't know i was supposed to be using those to get my answer.

- 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

no no i meant...

- anonymous

cos^2 - 1

- anonymous

that's part of cos^2 + sin^2 = 1 right?

- anonymous

you used that to get the answer?

- 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

okay thanks! off to take my test.

- anonymous

Have fun!

- anonymous

lol sure

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