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Trignometric Identities Problem

  1. Jan 29, 2005 #1
    A little confused on how to begin the problem


    _1__ - tanx sinx = cosx
    cosx

    I know you change tanx to sinx/cosx but I cant seem to finish the problem, not sure if it is arithmatic errors or what?
     
    Last edited: Jan 29, 2005
  2. jcsd
  3. Jan 29, 2005 #2
    Just simplify both sides, by taking the most logical step... And remember [itex]cos^2 x +sin^2 x=1[/itex]
     
  4. Jan 29, 2005 #3

    dextercioby

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    Homework Helper

    Write everything uder the same denominator.And the definition of tangent and the sine^{2}+cos^{2} connection.

    Daniel.
     
  5. Jan 29, 2005 #4
    You had the right idea starting with [tex] tanx = \frac{sinx}{cosx} [/tex]

    Are you familiar with the trig identity: [tex] sin^2 x + cos^2 x = 1 [/tex]. In many trig question, you have to multiply/divide out parts of your equation to work out the final solution. Take a look at the question and see if there are any common values that would be worth taking out.
     
  6. Jan 29, 2005 #5
    Alright so

    _1__ - tanx sinx = cosx
    cosx
    _1__ - sinx sinx = cosx
    cosx cosx
    _1__ - sinx
    cosx cosx

    Or is it

    cosx-cosxsinx * sinx
    ` 1```` 1 ```````1
     
  7. Jan 29, 2005 #6
    [tex]\frac{1}{cos\theta}-\frac{sin^2\theta}{cos\theta}=cos\theta[/tex]
    Now find a common denominator and simplify it.
     
  8. Jan 29, 2005 #7

    dextercioby

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    OMG,okay here goes
    [tex] \frac{1}{\cos x}-\frac{\sin x}{\cos x} \sin x=... [/tex]

    Can u take it from here?

    Daniel.
     
  9. Jan 30, 2005 #8
    ok, i guess ill provide a little more help:
    [tex]\frac{1}{cos\theta}-tan\theta sin\theta=cos\theta[/tex]

    simplify the tan:
    [tex]\frac{1}{cos\theta}-\frac{sin\theta}{cos\theta}sin\theta=cos\theta[/tex]

    multiply and subtract, because you have like denominators:
    [tex]\frac{1-sin^2\theta}{cos\theta}=cos\theta[/tex]

    now, use the fact that [itex]sin^2\theta + cos^2\theta = 1[/itex] to solve.
     
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