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anonymous

  • one year ago

Verify the identity. cotx minus pi divided by two. = -tan x

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  1. misty1212
    • one year ago
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    HI!!

  2. misty1212
    • one year ago
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    if you can use the cofunction identity this is real easy

  3. misty1212
    • one year ago
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    since cotangnent is even, you know \[\cot(x-\frac{\pi}{2})=-\cos(\frac{\pi}{2}-x)\]

  4. misty1212
    • one year ago
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    damn i meant "odd"

  5. misty1212
    • one year ago
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    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

  6. anonymous
    • one year ago
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    yea, its tan x right?

  7. misty1212
    • one year ago
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    damn i am making a lot of typos

  8. misty1212
    • one year ago
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    let me try again in one line \[\cot(x-\frac{\pi}{2})=-\cot(\frac{\pi}{2}-x)=-\tan(x)\]

  9. anonymous
    • one year ago
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    yea, -tan x right?

  10. misty1212
    • one year ago
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    the first equal sign because cotangent is odd the second equal sign in the cofunction identity

  11. misty1212
    • one year ago
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    is this clear? it is really only two steps, but maybe they are not obvious

  12. anonymous
    • one year ago
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    the identity is -tan x or am I lost.

  13. anonymous
    • one year ago
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    Is it cos(pi/2-x) ?

  14. misty1212
    • one year ago
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    i thought that might be the case lets go slow and take them one at a time the cofunction identities say \[\cos(\frac{\pi}{2}-x)=\sin(x)\\ \csc(\frac{\pi}{2}-x)=\sec(x)\\ \cot(\frac{\pi}{2}-x)=\tan(x)\]

  15. anonymous
    • one year ago
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    cot(pi/2-x)

  16. misty1212
    • one year ago
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    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

  17. misty1212
    • one year ago
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    is that part clear?

  18. anonymous
    • one year ago
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    Oh ok, yes it is. Thnx!! :)

  19. misty1212
    • one year ago
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    ok and how about the first part \[\cot(x-\frac{\pi}{2})=-\cot(\frac{\pi}{2}-x)\]?

  20. anonymous
    • one year ago
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    yea, that's the part that kinda got me.

  21. misty1212
    • one year ago
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    ok lets do that slow too

  22. anonymous
    • one year ago
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    Ok :)

  23. misty1212
    • one year ago
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    cotangent in an "odd" function, which means \[\cot(-x)=-\cot(x)\] in general "odd" means \[f(-x)=-f(x)\]

  24. misty1212
    • one year ago
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    sine and tangent are also "odd" so \[\sin(-x)=-\sin(x)\] etc

  25. misty1212
    • one year ago
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    now \[-(x-\frac{\pi}{2})=\frac{\pi}{2}-x\] right ?

  26. anonymous
    • one year ago
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    Yes...

  27. misty1212
    • one year ago
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    that means, since cotangent is odd, that \[\cot(x-\frac{\pi}{2})=\cot(\color{red}-(\frac{\pi}{2}-x))=\color{red}-\cos(\frac{\pi}{2}-x)\]

  28. misty1212
    • one year ago
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    notice the minus sign comes out of the function

  29. anonymous
    • one year ago
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    oh yea, this makes sense....

  30. misty1212
    • one year ago
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    it is a bit abstract, but not too hard just realizing that \(a-b=-(b-a)\) and therefore \(\cot(a-b)=-\cos(b-a)\)

  31. anonymous
    • one year ago
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    oh, wow it's so much clearer now. Thank you @misty1212 for your help.

  32. misty1212
    • one year ago
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    \[\huge \color\magenta\heartsuit\]

  33. anonymous
    • one year ago
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    <3

  34. anonymous
    • one year ago
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    Lol

  35. misty1212
    • one year ago
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    thanks

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