1. ## trigonometry problem

using the substitution x= tan a, show that

y = tan-1 2x /1+x2

reduces to

y = 2 tan-1 x

this is what i have done:-

tan y = 2x/1+x2

tan y = 2 tan a / 1 + tan2 a

tan y = 2 tan a / sec2 a

tan y = 2sin a cos a = sin 2a

i cant work this out further...can you help me plz?

2. Originally Posted by slash
using the substitution x= tan a, show that

y = tan-1 2x /1+x2

reduces to

y = 2 tan-1 x

this is what i have done:-

tan y = 2x/1+x2

tan y = 2 tan a / 1 + tan2 a

tan y = 2 tan a / sec2 a

tan y = 2sin a cos a = sin 2a

i cant work this out further...can you help me plz?
do you mean: $\displaystyle y = \tan^{-1} \left( \frac {2x}{1 + x^2} \right)$ ?

3. Originally Posted by Jhevon
do you mean: $\displaystyle y = \tan^{-1} \left( \frac {2x}{1 + x^2} \right)$ ?

yes ur right..
sorry im new to all this..i dont know how to use em..

4. Originally Posted by slash
yes ur right..
sorry im new to all this..i dont know how to use em..
first off, i think the denominator should be: 1 - x^2

secondly, you would write it like this: y = arctan [(2x)/(1 - x^2)] .......arctan means the inverse tangent. or you could write tan^{-1} [(2x)/(1 - x^2)]

Now here's how we do it:

$\displaystyle y = \tan^{-1} \left( \frac {2x}{1 - x^2} \right)$

$\displaystyle \Rightarrow \tan y = \frac {2x}{1 - x^2}$

Let $\displaystyle x = \tan a$

$\displaystyle \Rightarrow \tan y = \frac {2 \tan a }{ 1 - \tan^2 a}$

But the right hand side is the double angle formula for tangent, so we get:

$\displaystyle \tan y = \tan 2a$

$\displaystyle \Rightarrow y = \tan^{-1} ( \tan 2a ) = 2a$

Since, $\displaystyle x = \tan a \implies a = \tan^{-1} x$, we have:

$\displaystyle y = 2 \tan^{-1} x$

as desired

5. Originally Posted by slash
using the substitution x= tan a, show that

y = tan-1 2x /1+x2

reduces to

y = 2 tan-1 x

this is what i have done:-

tan y = 2x/1+x2

tan y = 2 tan a / 1 + tan2 a

tan y = 2 tan a / sec2 a

tan y = 2sin a cos a = sin 2a

i cant work this out further...can you help me plz?
You converted [1 +tan^2(a)] into sec^2(a). Although the conversion is true, it is not needed here.

2tan(a) / 1+tan^2(a)
is tan(2a).
So,
tan(y) = tan(2a)
Get the arctans both sides,
y = 2a -----------------------(i)

But x = tan(a)
So, getting the arctans of both sides,
arctan(x) = a
Substitute that into (i),
y = 2[arctan(x)] --------------shown.

6. Hello, slash!

I agree with Jhevon . . . there is a typo.

Using the substitution $\displaystyle x\,= \,\tan a,$

show that: .$\displaystyle y \:= \:\tan^{-1}\left(\frac{2x}{1-x^2}\right)$ reduces to: .$\displaystyle y \:= \:2\tan^{-1}(x)$
. . . . . . . . . . . . . . . . . .$\displaystyle {\color{red}\uparrow}$

We're expected to know this identity: .$\displaystyle \tan2\theta \:=\:\frac{2\tan\theta}{1-\tan^2\!\theta}$

Making that substitution, we have:

. . $\displaystyle y \;=\;\tan^{-1}\left(\frac{2\tan a}{1 - \tan^2\!a}\right) \;=\;\tan^{-1}(\tan2a) \;=\;2a$

Since $\displaystyle x = \tan a\!:\;\;a \,=\,\tan^{-1}(x)$

. . Therefore: .$\displaystyle y \;=\;2\tan^{-1}(x)$