Trigonometry help

[shawonda]

Hi. I was wondering if someone could help me on a problem that I have been working on for four days and I still can't figure it out. I can't use the Multiple Angle Identity, so if you can figure it out not using the double angle, multiple angle formulas I'd really appreciate the help.

Here goes the problem:
You have to verify the identity

cot2x=[(cot^2 x) -1]/(2*cotx

If I can get help figuring out this one then I can find out how to do the other one similar to it.

Thanks in advance...shawonda

 

[stapel]

...a problem that I have been working on for four days....

There are probably many ways to approach this exercise. Please reply showing some of the avenues you've tried, or at least the approach on which you feel you've made the most progress.

Thank you.

Eliz.

 

[skeeter]

working from the right side, start by multiplying both the numerator and denominator by sin²x ...

sin²x[cot²x - 1]/[sin²x(2cotx)] =

[cos²x - sin²x]/[2sinxcosx] =

cos(2x)/sin(2x) = cot(2x)

you're done.

 

[shawonda]

Yeah, I tried the multiplication by sin^2x but then for the life of me couldn't figure out how to get to cos2x/sin2x without using the double angle formula or the half angle formula. I have already done it that way but doing it without the two formulas mentioned before has me stumped. I have tried multiplication by 1/cotx but then that just gets me (cos^2x-sin^2x)/(2cosx*sinx) and then I'm stuck all over again. I have tried multiplication by sinx/cosx and that got me (cotx -1)/2. Is there a way to do it without using double angle or half angle identities?

 

[shawonda]

Thanks for replying to my post. I figured out the problem by working with the cot2x side instead of the cot^xx/1/2cotx side. I was able to get the answer by working with the reciprocal of cot2x.

Thank you,

shawonda