Half angle identities?

Question

22.5degrees=tan(45/2)

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Answer 1 · 2018-03-26T03:23:52

$tan22.5^@=sqrt2-1$

Here are the $sin$ , $cos$ , and $tan$ half-angle identities:

$sin(theta/2)=+-sqrt((1-costheta)/2)$

$cos(theta/2)=+-sqrt((1+costheta)/2)$

$tan(theta/2)=+-sqrt((1-costheta)/(1+costheta))$

$color(white)(tan(theta/2))=sintheta/(1+costheta)$

$color(white)(tan(theta/2))=costheta/(1-sintheta)$

To figure out the $tan$ of $22.5^@$ , use the $tan$ half-angle formula. I personally like the second one the best, so I'll use that one:

$color(white)=tan(22.5^@)$

$=tan(45^@/2)$

$=sin(45^@)/(1+cos(45^@))$

$=(sqrt2/2)/(1+sqrt2/2)$

$=sqrt2/(2+sqrt2)$

$=((sqrt2))/((2+sqrt2))color(red)(*((2-sqrt2))/((2-sqrt2)))$

$=((sqrt2)(2-sqrt2))/((2+sqrt2)(2-sqrt2))$

$=(2sqrt2-sqrt2^2)/((2+sqrt2)(2-sqrt2))$

$=(2sqrt2-2)/((2+sqrt2)(2-sqrt2))$

$=(2sqrt2-2)/(4+2sqrt2-2sqrt2-sqrt2^2)$

$=(2sqrt2-2)/(4color(red)cancelcolor(black)(color(black)+2sqrt2-2sqrt2)-sqrt2^2)$

$=(2sqrt2-2)/(4-sqrt2^2)$

$=(2sqrt2-2)/(4-2)$

$=(2sqrt2-2)/2$

$=(color(red)cancelcolor(black)2sqrt2-color(red)cancelcolor(black)2^1)/color(red)cancelcolor(black)2$

$=sqrt2-1$

This is the result. You can check your answer using a calculator:

![https://www.desmos.com/calculator]( useruploads.socratic.org )

Hope this helped!