# double-angle formula

• Nov 26th 2008, 03:48 AM
jvignacio
double-angle formula
just wondering how this equals 4.

$\displaystyle \frac{2sin2xcos2x}{(cos2x)\frac{1}{2}sin2x} = 4$

once again abit confused why that equals 4. thank u.
• Nov 26th 2008, 04:16 AM
Soroban
Hello, jvignacio!

You need to brush up on basic fractions . . .

. . $\displaystyle \frac{2(\sin2x)(\cos2x)}{(\cos2x)(\frac{1}{2})(\si n2x)} \;=\;\frac{2({\color{blue}\rlap{/////}}\sin2x)({\color{red}\rlap{/////}}\cos2x)}{({\color{red}\rlap{/////}}\cos2x)(\frac{1}{2})({\color{blue}\rlap{/////}}\sin2x)}$ .$\displaystyle =\;\frac{2}{\frac{1}{2}} \;=\;4$

• Nov 26th 2008, 04:39 AM
jvignacio
Quote:

Originally Posted by Soroban
Hello, jvignacio!

You need to brush up on basic fractions . . .

. . $\displaystyle \frac{2(\sin2x)(\cos2x)}{(\cos2x)(\frac{1}{2})(\si n2x)} \;=\;\frac{2({\color{blue}\rlap{/////}}\sin2x)({\color{red}\rlap{/////}}\cos2x)}{({\color{red}\rlap{/////}}\cos2x)(\frac{1}{2})({\color{blue}\rlap{/////}}\sin2x)}$ .$\displaystyle =\;\frac{2}{\frac{1}{2}} \;=\;4$

ahh crap i forgot to cancel out . thanks man