#1
Sup Bitches?


It goes like this cot (x) * sin(x) = cos(x)


no numbers involved just switching shit around to make cos(x) = cos(x)

i got


                      1          *    sin(x)  =  cos(x)
                   --------          -------
                    sin(x)              1
                   --------
                    cos(x)



How do i deal with that fraction within a fraction bitch?


Thanks
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#5
i just did these, u in 12?

the cot can be turned into Cos^2/Sin^2

then you have

cos^2 Sin
--------- X --------
sin^2 Sin


= cos^2sin
------------
sin^2

then you factor the top, cross out the sins and you are done. i think, idk this stuff gave me trouble.
#7
In regards to the actual question you had to answer, if you still need help on it, once you have cos/sin you multiply that by sin giving you sincos/sin which allows you to cancel out the sins giving you cos
#8
cotangent = 1/(sin/cos) which is the exact same thing as (cos/sin)

remember that (2/3)/(3/4) is the same as (2/3) x (4/3), same logic applies here

(cos/sin)x(sin)=cos because the sines cancel out and you are left with the cosine.


Speaking of trig identities, I have a quiz on verifying identities and the like tomorrow in my precalc class which I should be studying for. Thanks for the reminder.

DivinEdit: And don't listen to anybody who tells you to use the Pythagorean identity. That is just over-complicating the problem.
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Last edited by divinecrossfire at Nov 23, 2009,
#9
cant cancle when mutiplied i dont think, and its not sin/sin= 1 its sin^2 + Cos^2 = 1

you will have to factor when you multiply then cancle
#10
Quote by misfitsramones
cant cancle when mutiplied i dont think, and its not sin/sin= 1 its sin^2 + Cos^2 = 1

you will have to factor when you multiply then cancle

No, you don't. Cot=cos/sin = 1/tan =1/sin/cos.
Cot*sin=Cos/sin * sin/1 = sincos/sin.

The sines divide out, and you're left with cos.
#11
cotx*sinx = (cosx/sinx)*sinx = cosx
Come on, TS, seriously...

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