Cosine A

magentarita · #1 $Old$ 08-29-2008, 04:48 AM

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.

mr fantastic · #2 $Old$ 08-29-2008, 04:59 AM

Quote:

Originally Posted by magentarita $View Post$

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.

$\cos A = \cos \left( 2 \left[ \frac{A}{2} \right] \right) = 2 \cos^2 \left( \frac{A}{2} \right) - 1$ .

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

Therefore $\cos A = \frac{2}{z^2 + 1} - 1 = ....$

nikhil · #3 $Old$ 08-29-2008, 05:04 AM

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

magentarita · #4 $Old$ 08-29-2008, 06:14 AM

Quote:

Originally Posted by nikhil $View Post$

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

A much better reply and easier to follow.