anonymous
  • anonymous
Verify the identity. cotx minus pi divided by two. = -tan x
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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misty1212
  • misty1212
HI!!
misty1212
  • misty1212
if you can use the cofunction identity this is real easy
misty1212
  • misty1212
since cotangnent is even, you know \[\cot(x-\frac{\pi}{2})=-\cos(\frac{\pi}{2}-x)\]

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More answers

misty1212
  • misty1212
damn i meant "odd"
misty1212
  • misty1212
in any case , as cotangent is odd, you have \[\cot(x-\frac{\pi}{2})=-\cos(\frac{\pi}{2}-x)\] and the cofunction identity tells gives \[\cos(\frac{\pi}{2}-x)=\tan(x)\] and you are done
anonymous
  • anonymous
yea, its tan x right?
misty1212
  • misty1212
damn i am making a lot of typos
misty1212
  • misty1212
let me try again in one line \[\cot(x-\frac{\pi}{2})=-\cot(\frac{\pi}{2}-x)=-\tan(x)\]
anonymous
  • anonymous
yea, -tan x right?
misty1212
  • misty1212
the first equal sign because cotangent is odd the second equal sign in the cofunction identity
misty1212
  • misty1212
is this clear? it is really only two steps, but maybe they are not obvious
anonymous
  • anonymous
the identity is -tan x or am I lost.
anonymous
  • anonymous
Is it cos(pi/2-x) ?
misty1212
  • misty1212
i thought that might be the case lets go slow and take them one at a time the cofunction identities say \[\cos(\frac{\pi}{2}-x)=\sin(x)\\ \csc(\frac{\pi}{2}-x)=\sec(x)\\ \cot(\frac{\pi}{2}-x)=\tan(x)\]
anonymous
  • anonymous
cot(pi/2-x)
misty1212
  • misty1212
since \[\cot(\frac{\pi}{2}-x)=\tan(x)\] that is the same as \[-\cot(\frac{\pi}{2}-x)=-\tan(x)\] just change the sign of both sides
misty1212
  • misty1212
is that part clear?
anonymous
  • anonymous
Oh ok, yes it is. Thnx!! :)
misty1212
  • misty1212
ok and how about the first part \[\cot(x-\frac{\pi}{2})=-\cot(\frac{\pi}{2}-x)\]?
anonymous
  • anonymous
yea, that's the part that kinda got me.
misty1212
  • misty1212
ok lets do that slow too
anonymous
  • anonymous
Ok :)
misty1212
  • misty1212
cotangent in an "odd" function, which means \[\cot(-x)=-\cot(x)\] in general "odd" means \[f(-x)=-f(x)\]
misty1212
  • misty1212
sine and tangent are also "odd" so \[\sin(-x)=-\sin(x)\] etc
misty1212
  • misty1212
now \[-(x-\frac{\pi}{2})=\frac{\pi}{2}-x\] right ?
anonymous
  • anonymous
Yes...
misty1212
  • misty1212
that means, since cotangent is odd, that \[\cot(x-\frac{\pi}{2})=\cot(\color{red}-(\frac{\pi}{2}-x))=\color{red}-\cos(\frac{\pi}{2}-x)\]
misty1212
  • misty1212
notice the minus sign comes out of the function
anonymous
  • anonymous
oh yea, this makes sense....
misty1212
  • misty1212
it is a bit abstract, but not too hard just realizing that \(a-b=-(b-a)\) and therefore \(\cot(a-b)=-\cos(b-a)\)
anonymous
  • anonymous
oh, wow it's so much clearer now. Thank you @misty1212 for your help.
misty1212
  • misty1212
\[\huge \color\magenta\heartsuit\]
anonymous
  • anonymous
<3
anonymous
  • anonymous
Lol
misty1212
  • misty1212
thanks

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