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Homework Help: Trig Identities : Help 3 Questions.

  1. Dec 12, 2004 #1
    I can't get these 3 questions, can someone help me?

    1. cotB [ (tanA + TanB) / (cotA+cotB) ] = tan A

    2. (sin^2A + 2cosA - 1) / (2 + cosA - cos^2A) = 1 / (1+ secA)

    3. cos^3A + sin^3A = (cosA+SinA)(1-SinAcosA)

    please help out on these, thx in advance. U can write the / sign as fractions cuz i can't do it on the computer here.
     
  2. jcsd
  3. Dec 12, 2004 #2

    Hurkyl

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    For proving trig identities, it's generally useful to convert everything into sin and cos, cross-multiply all fractions, and multiply out all factorizations.
     
  4. Dec 12, 2004 #3

    Diane_

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    Homework Helper

    It's also helpful to know the double-angle formulas and the half-angle formulas.
     
  5. Dec 12, 2004 #4
    no need to use double-angle formulas for those... i just can't figure them out. can someone actually do one?
     
  6. Dec 12, 2004 #5

    kreil

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    here is the first one:
    [tex]1. tan(A)=cot(B)\frac{tan(A) + tan(B)}{cot(A)+cot(B)}[/tex]

    Express in terms of sine and cosine:
    [tex]=(\frac{\cos{B}}{\sin{B}})\frac{\frac{\sin{A}}{\cos{A}}+\frac{\sin{B}}{\cos{B}}}{\frac{cosA}{sinA}+\frac{cosB}{sinB}}[/tex]

    Get common denominators and add the top/bottom to form 1 complex fraction:
    [tex]=(\frac{\cos{B}}{\sin{B}})\frac{\frac{sinAcosB+sinBcosA}{cosAcosB}}{\frac{sinBcosA+sinAcosB}{sinAsinB}}[/tex]

    Simplify:
    [tex]=(\frac{\cos{B}}{\sin{B}})\frac{\frac{sin(A+B)}{cosAcosB}}{\frac{sin(A+B)}{sinAsinB}}[/tex]

    [tex]=\frac{\frac{sin(A+B)}{cosA}}{\frac{sin(A+B)}{sinA}}=\frac{sinA}{cosA}=tanA[/tex]

    Identities used in solution:

    [tex]cotA=\frac{cosA}{sinA}[/tex]

    [tex]tanA=\frac{sinA}{cosA}[/tex]

    [tex]sin(A+B)=sinAcosB+cosAsinB[/tex]

    Good luck with the others, I hope this helps you!
     
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