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there is something wrong in the expression!!
the tan-1 is inverse tan not minus 1
oh I see
sorry about that
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?
Maybe it makes more sense like this: let α = arctan(x) <=> tan(α) = x Therefore the opposite side over the ajacent side = .... ,
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 = ....
sorry i was grabbing lunch
OK then. Do you see where I'm going, though?
yeah i understand. a^2 +b^2 = c ^2. what i don't understand is where the 2 comes int0 play
before the tangent
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)
im sorry but that doesn't make any sense to me. my brain is peicing together what you're saying
Does that image help explain it better?
hold on one second
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)
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)
so if i just take what i find for tan (opposite/adjacent) and distribute a 2 before putting it back into terms of cosine?
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
You know tan(a) = x by definition
okay so cos(2a) got it. so you're just using tan to find the value's of the triangle?
Yes, from which you can work out cos (or sin, but you don't need both here)
ohhhhh okay. tan-1x = a ? so 2tan-1x = 2a right then cos(2a) is what i want?
Yes, indeed :D
ahh thanks let me work it out and ill reply with an answer
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).
my book says the answer is (1-x^2)/(1+x^2)
but idk how to get there
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
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)
Sorry the simplification/fraction look messy but hopefully the brackets make it clear
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.
those are pythag identities ?
thats what i was missing. i didin't know i was supposed to be using those to get my answer.
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
no no i meant...
cos^2 - 1
that's part of cos^2 + sin^2 = 1 right?
you used that to get the answer?
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.
okay thanks! off to take my test.