Question

HELP

Answer 1

Show that \[\frac{ d }{ dx }\left( \tan \frac{ x }{ 2 }\right)=\frac{ 1 }{ 1+cosx }\]

Answer 2

Who can help me?

Answer 3

Give me a sec.

Answer 4

We express the tan(x/2) as sin(x/2) / cos (x/2).

d/dx of tan(x/2) = d/dx (sin(x/2) / cos (x/2)) =

( sin(x/2) ' * cos(x/2) - sin(x/2)*cos(x/2) ' ) / ( cos(x/2) ) ^2 <=>

(cox(x/2)^2*1/2 + sin(x/2)^2*1/2) / ( cos(x/2)^2 ) <=>

1/2* (sin(x/2)^2 + cos(x/2)^2) / ( cos(x/2)^2 ) <=>

1/ ( 2* ( cos(x/2)^2 ). (1)

***Now, we need to express cox(x/2) as something times cox(x). The simplest way to do this without using any other predefined formulas is to write:

cox(x)=cox(x/2 + x/2)= cox(x/2)^2 - sin(x/2)^2. (2)

Fundamental identity : cos(x/2)^2 + sin(x/2)^2 = 1. (3)

We add (2) and (3)=> cos(x) + 1= cox(x/2)^2 - sin(x/2)^2.+ cos(x/2)^2 + sin(x/2)^2

= 2*cos(x/2)^2. (4)

Which perfectly fits our initial expression (1), thus:

d/dx of tan(x/2) = 1/ ( 2* ( cos(x/2)^2 ) =1/( cox(x) +1). ( according to (4) )

*** We could bypass (4) by simply using a trigonometric identity : cos(2x) = cos(x)^2 - sin(x)^2 = 2*cos(x)^2 - 1 which is supposedly known but then it wouldn't be as fun.

Answer 5

I'm sorry for the long post but I went along trigonometry striving not to memorize identities but rather knowing how to approach them with a fundamental set of identities. If you have any questions, feel free to ask.