# trigonometry problem

• Aug 26th 2007, 07:55 AM
slash
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?
• Aug 26th 2007, 08:02 AM
Jhevon
Quote:

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

• Aug 26th 2007, 08:08 AM
slash
Quote:

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..
• Aug 26th 2007, 08:19 AM
Jhevon
Quote:

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
• Aug 26th 2007, 12:05 PM
ticbol
Quote:

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.
• Aug 26th 2007, 01:29 PM
Soroban
Hello, slash!

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

Quote:

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