1. ## help please cos x

By writing cos x in terms of 1/2x show that

1 - cosx / 1 + cosx = tan^2 1/2x

thankyou :0

2. Originally Posted by gracey
By writing cos x in terms of 1/2x show that

1 - cosx / 1 + cosx = tan^2 1/2x

thankyou :0
$\frac{1 - \cos\left[ 2 \left (\frac{x}{2} \right] \right)}{1 + \cos \left (2 \left[ \frac{x}{2} \right] \right)} = \frac{1 - \cos^2 \left (\frac{x}{2}\right) + \sin^2 \left (\frac{x}{2}\right)}{1 + \cos^2 \left (\frac{x}{2}\right) - \sin^2 \left (\frac{x}{2}\right)} = \frac{2\sin^2 \left (\frac{x}{2}\right)}{2 \cos^2 \left (\frac{x}{2}\right)} = \tan^2 \left (\frac{x}{2}\right)$

where the Pythagorean Identity $\sin^2 \left ( \frac{x}{2} \right) + \cos^2 \left ( \frac{x}{2} \right) = 1$ has been used in the numerator and the denominator.

3. Hello, gracey!

A slightly different approach . . .

We have these identities:

. . $\begin{array}{ccccccc}\sin^2\!\frac{x}{2} &=& \dfrac{1-\cos x}{2} & \Rightarrow & 1-\cos x &=&2\sin^2\!\frac{x}{2} \\ \\[-4mm]
\cos^2\!\frac{x}{2} &=&\dfrac{1+\cos x}{2} & \Rightarrow & 1 + \cos x &=& 2\cos^2\!\frac{x}{2} \end{array}$

By writing $\cos x$ in terms of $\frac{x}{2}$ show that: . $\frac{1 - \cos x}{1 + \cos x} \:= \:\tan^2\!\frac{x}{2}$

From the identities, we have: . $\frac{1-\cos x}{1 + \cos x} \;=\;\frac{2\sin^2\!\frac{x}{2}}{2\cos^2\!\frac{x} {2}} \;=\;\left(\frac{\sin\frac{x}{2}}{\cos\frac{x}{2}} \right)^2 \;=\;\tan^2\!\frac{x}{2}$

4. ## Here is what you need!

Hi!gracey,hope ur having a great time!
cosx=(cos(x/2))^2-(sin(x/2))^2
since (cos(x/2))^2+(sin(x/2))^2=1
and A=A/1
so cosx=[(cos(x/2))^2-(sin(x/2))^2]/1
=[(cos(x/2))^2-(sin(x/2))^2]/
[(cos(x/2))^2+(sin(x/2))]
dividing numerator and denominator by (cos(x/2))^2 we get
cosx=[1-(tan(x/2))^2]/[1+(tan(x/2))^2]
now cross multiply
[1+(tan(x/2))^2]cosx=1-(tan(x/2))^2
cosx+cosx(tan(x/2))^2=1-(tan(x/2))^2
cosx(tan(x/2))^2+(tan(x/2))^2=1-cosx
[(tan(x/2))^2][cosx+1]=1-cosx
(tan(x/2))^2=(1-cosx)/1+cosx
or
(1-cosx)/1+cosx=(tan(x/2))^2