Double Angles

**Sherina** · November 9th 2008, 01:09 PM

How would I simplify:

(tan22.5)/1 - tan^2(22.5)

And also:

let csc t = (square root)5 with t in QII and find sec 2t

I have no idea where to start

**Arch_Stanton** · November 10th 2008, 02:40 AM

$=\frac{\frac{sin x}{cos x}}{1-\frac{sin^2 x}{cos^2 x}}=...=\frac{sin x cos x}{cos^2 x-sin^2 x}=\frac{2sin x cos x}{2(cos^2 x-sin^2 x)}=\frac{sin2x}{2cos2x}=\frac{sin 45^o}{2cos 45^o}=\frac{1}{2}$

2. $sec=\frac{1}{cos x}$
$csc x=\frac{1}{sin x}$
$csc t=\sqrt{5} \Rightarrow sin t=\frac{\sqrt{5}}{5}$
from the pythagorean trigonometric identity:
$sin^2 x+cos^2 x=1$
$cos t=\sqrt{1-\frac{5}{25}}=\frac{2}{\sqrt{5}}$
$cos 2t=cos^2 t-sin^2 t=\frac{4}{5}-\frac{1}{5}=\frac{3}{5}$
$sec 2t=\frac{1}{cos 2t}=\frac{5}{3}$

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