Our teacher gave the class a challenge problem.

If z = tan(A/2), show that cos(A) = (1 - z^2)/(1 + z^2)

I know that replacing z^2 with tan(A/2) is a start but have no idea where to go from there.

Printable View

- Aug 29th 2008, 02:48 AMmagentaritaCosine A
Our teacher gave the class a challenge problem.

If z = tan(A/2), show that cos(A) = (1 - z^2)/(1 + z^2)

I know that replacing z^2 with tan(A/2) is a start but have no idea where to go from there.

- Aug 29th 2008, 02:59 AMmr fantastic
$\displaystyle \cos A = \cos \left( 2 \left[ \frac{A}{2} \right] \right) = 2 \cos^2 \left( \frac{A}{2} \right) - 1$.

$\displaystyle \tan \left( \frac{A}{2} \right) = z \Rightarrow \cos \left( \frac{A}{2} \right) = \frac{1}{\sqrt{z^2 + 1}}$.

Therefore $\displaystyle \cos A = \frac{2}{z^2 + 1} - 1 = .... $ - Aug 29th 2008, 03:04 AMnikhil
Hope you know the identity

cos2A=(cosA)^2-(sinA)^2 which is equal to

cos2A=[(cosA)^2-(sinA)^2]/1

but (cosA)^2+(sinA)^2=1 therfor

cos2A=[(cosA)^2-(sinA)^2]/[(cosA)^2+(sinA)^2]

now dividing both numerator and denominator by (cosA)^2

we get

cos2A=[(1-(tanA)^2]/[(1+(tanA)^2]

now put A/2 in place of A and you get

cosA=[(1-(tanA/2)^2]/[(1+(tanA/2)^2]

but tanA/2=z therfor

cosA=[(1-z^2]/[(1+z^2]

hence proved - Aug 29th 2008, 04:14 AMmagentaritaBetter...