Use double angle identity to find value of tan(x) / [1-...

[Guest]

Using the identity "tan (2a) = [ 2 tan(a) ] / [1 - tan^2(a) ], find the value of tan (x) / [1 - tan^2(x/2) ].

I derived this to be tan x ie...tan(x/2 + x/2). But when I do my working out I dont get tan (x) /1-tan^2(x/2)...

Is tan(x) the real answer?

thank you in advance

 

[stapel]

The only way ^tan(x) / _{(1 - tan^2(x/2))} could equal tan(x) would be for the denominator, 1 - tan²(x/2), to equal 1. Of course, this isn't true. So....

What did you "derive" to be which?

Please reply showing all of your work and reasoning. Thank you.

Eliz.

 

[soroban]

Re: Use double angle identity to find value of tan(x) / [1-.

Hello, americo74!

Using the identity: \(\displaystyle \:\tan (2a) \:=\:\L\frac{2\tan(a)}{1\,-\,\tan^2(a)}\)
. . find the value of: \(\displaystyle \L\:\frac{\tan(x)}{1\,-\,\tan^2\left(\frac{x}{2}\right)}\)

Since no numbers are given, what do they mean by "find the value"?


That half-angle is annoying . . . Let \(\displaystyle u\,=\,\frac{x}{2}\;\;\Rightarrow\;\;x\,=\,2u\)

Then we have: \(\displaystyle \L\:\frac{\tan(2u)}{1\,-\,\tan^2(u)}\)

Using that identity, the numerator is: \(\displaystyle \:\tan(2u)\:=\:\L\frac{2\tan)u_}{1\,-\,\tan^2(u)}\)

And the problem becomes: \(\displaystyle \L\:\frac{\frac{2\tan(u)}{1\,-\,\tan^2(u)} }{1\,-\,\tan^2(u)} \:=\:\frac{2\tan(u)}{\left[1\,-\,\tan^2(u)\right]^2}\)

Replace \(\displaystyle u\) with \(\displaystyle \frac{x}{2}:\;\;\L\frac{2\tan\(\frac{x}{2})}{\left[1\,-\,\tan^2(\frac{x}{2})\right]^2}\;\;\) . . . and that's all we can do

 

[Guest]

find the value means

use the identity to simplify the expression that is all..

 

[Count Iblis]

I would prefer to simplify it differently (but this is a subjective thing...)


\(\displaystyle \frac{\tan(x)}{1-\tan^{2}(\frac{x}{2})}=\frac{2\tan(\frac{x}{2})}{1-\tan^{2}(\frac{x}{2})}\times\frac{\tan(x)}{2\tan(\frac{x}{2})}=\frac{\tan^{2}(x)}{2\tan(\frac{x}{2})}\)

The advantage of such a form, in which things are expressed as products or quotients of trignometric functions, is that you can immediately read-off the behavior of the function.