1. Trigonometric Equations

I'm not sure how to tackle this question, can anyone help me out? It asks;

a) Show that $\displaystyle \frac{1 - cos2x}{1 + cos2x} = tan^2x$.

b) Hence find the value of $\displaystyle tan22.5^o$ in simplest exact form.

2. Originally Posted by Flay
I'm not sure how to tackle this question, can anyone help me out? It asks;

a) Show that $\displaystyle \frac{1 - cos2x}{1 + cos2x} = tan^2x$.

b) Hence find the value of $\displaystyle tan22.5^o$ in simplest exact form.
$\displaystyle \frac{1 - cos2x}{1 + cos2x} = \frac{1 - (cos^2x - sin^2x)}{1+cos^2x-sin^2x}=\frac{1-cos^2x+sin^2x}{cos^2x+cos^2x}$$\displaystyle =\frac{sin^2x+sin^2x}{2cos^2x}=\frac{2sin^2x}{2cos ^2x}=tan^2x$

For b, take the square root of both sides and you get an equation for $\displaystyle tan(x)$, let x = 22.5.

3. Originally Posted by Flay
I'm not sure how to tackle this question, can anyone help me out? It asks;

a) Show that $\displaystyle \frac{1 - cos2x}{1 + cos2x} = tan^2x$.

b) Hence find the value of $\displaystyle tan22.5^o$ in simplest exact form.
For part a, focus on the left side of the equation.

First multiply the numerator and denominator by $\displaystyle \tfrac{1}{2}$

$\displaystyle \frac{\tfrac{1}{2}(1-\cos(2x))}{\tfrac{1}{2}(1+\cos(2x))}$

Note that $\displaystyle \cos^2(x)=\tfrac{1}{2}(1+\cos(2x))$ and $\displaystyle \sin^2(x)=\tfrac{1}{2}(1-\cos(2x))$

Thus, the identity becomes $\displaystyle \frac{\sin^2(x)}{\cos^2(x)}=\tan^2(x)$. We have proven the identity. Does this make sense?

For part (b), note that $\displaystyle \tan(22.5^{\circ})=\tan\left(\tfrac{45^{\circ}}{2} \right)$

Apply the half angle identity $\displaystyle \tan\left(\tfrac{\vartheta}{2}\right)=\frac{\sin\l eft(\tfrac{\vartheta}{2}\right)}{\cos\left(\frac{\ vartheta}{2}\right)}=\sqrt{\frac{1-\cos\vartheta}{1+\cos\vartheta}}$, where $\displaystyle \vartheta=45^{\circ}$.

Does this make sense?

--Chris

4. Thanks, I understand now. I was making a mistake with my postive and negative signs in the simplification of the first equation.