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July 23, 2014

July 23, 2014

Posted by **Gina** on Tuesday, May 27, 2008 at 7:35pm.

cot(x/4) - cot(x) = [sin(kx)]/[sin(x/4)sin(x)]

PLEASEEEE HELP! ASAP!

- Math -
**???**, Tuesday, May 27, 2008 at 8:54pmThat is a very confusing question, you may want to reword that if you want an answer

- Math -
**Reiny**, Tuesday, May 27, 2008 at 10:19pmthe left side

= cot(x/4) - cot(x)

= [cos(x/4)/sin(x/4)] - cos(x)/sin(x)

= [sin(x)cos(x/4)- cos(x)sin(x/4)]/[sin(x/4)sin(x)]

= sin(x - x/4)/[sin(x/4)sin(x)]

comparing left side with right side, we notice the denominators are the same, so the numerator has to be the same

then sin(x - x/4) = sin (kx)

and x-x/4 = kx

3x/4 = kx

k = 3/4

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