From: http://en.wikipedia.org/wiki/Tangent_half-angle_substitution

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How did $\frac 1 {2\cos ^2 \frac x 2}$ become: $\frac {1+t^2} 2$?

From the substitution of $\cos x$, it should be similar to: $\frac 1 {\cos^2x} = \frac {t^2+1}{1-t^2}$ no?

I'm asking this because I'm trying to make this substitution for $\sin 2x, \cos 2x$, using $t=\tan 2x$ (and they turn out to be the same as the ones for $t=\tan \frac x 2$) but I can't find the $dx$.

  • 2
    $\begingroup$ We have $\sec^2 w=\tan^2 w+1$. If this identity is not familiar, note that $1=\sin^2 w+\cos^2 w$ and divide both sides by $\cos^2 w$. $\endgroup$ – André Nicolas Apr 24 '15 at 14:36
  • $\begingroup$ Ah I see. So all of those 3 substitutions (for $\sin x ,\cos x, dx $) would be the same for any $w$ in $t=tan w$? @AndréNicolas $\endgroup$ – shinzou Apr 24 '15 at 14:44
  • $\begingroup$ I don't know what you mean. The substitutions for $\sin x$, $\cos x$, and $dx$ all use trigonometric identities, but they are different identities. $\endgroup$ – André Nicolas Apr 24 '15 at 15:03

The $t$ in these substitutions is $t=\tan(x/2)$, not $t=\tan x$ so that $$\sec^2(x/2)= \tan^2 (x/2)+1=t^2+1$$


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