Question

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

Answer 1

there is something wrong in the expression!!

Answer 2

the tan-1 is inverse tan not minus 1

Answer 3

oh I see

Answer 4

sorry about that

Answer 5

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?

Answer 6

Maybe it makes more sense like this:

let α = arctan(x) <=> tan(α) = x

Therefore the opposite side over the ajacent side = .... ,

Answer 7

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 = ....

Answer 8

sorry i was grabbing lunch

Answer 9

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

Answer 10

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

Answer 11

before the tangent

Answer 12

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)

Answer 13

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

Answer 14

piecing*

Answer 15

Does that image help explain it better?

Answer 16

hold on one second

Answer 17

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)

Answer 18

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)

Answer 19

so if i just take what i find for tan (opposite/adjacent) and distribute a 2 before putting it back into terms of cosine?

Answer 20

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

Answer 21

You know tan(a) = x by definition

Answer 22

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

Answer 23

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

Answer 24

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

Answer 25

Yes, indeed :D

Answer 26

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

Answer 27

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

Answer 28

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).

Answer 29

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

Answer 30

but idk how to get there

Answer 31

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

Answer 32

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)

Answer 33

okay yes

Answer 34

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

Answer 35

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.

Answer 36

those are pythag identities ?

Answer 37

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

Answer 38

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

Answer 39

no no i meant...

Answer 40

cos^2 - 1

Answer 41

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

Answer 42

you used that to get the answer?

Answer 43

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.

Answer 44

okay thanks! off to take my test.

Answer 45

Have fun!

Answer 46

lol sure