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Simplifying trig functions

  1. Jul 31, 2003 #1
    I am currently learning how to simplify trig functions, but is there a way to know which formulae to use?

    In my text book there are three formulae:

    cos^2ttheta + sin^2theta = 1
    1 + tan^2theta = sec^2theta
    cot^2theta + 1 = cosec^2theta

    I am also stuck in this question:

    prove this identity:

    cosA / (1 - tanA) + sinA / (1 - cotA) = Sin A + cosA

    Since this requires a proof I need to work on the LHS and RHS separately. Can someone help me start this question? I really appreciate your help.
     
  2. jcsd
  3. Jul 31, 2003 #2
    First off, work on the complicated side, which is the LHS.

    Usually if we see tan A and cot A in an expression together with sin A and cos A, we change tan A to sin A/cos A and cot A to cos A/ sin A before further simplifying the expression. In this question, you don't need to use the 3 formulae listed.
     
  4. Jul 31, 2003 #3
    well, the best I have done is this stage:

    (cos^2A - sin^2A / cosA) + (sin^2A - cos^2A / sinA)
     
  5. Jul 31, 2003 #4
    Please show your working as I think the expression you got isn't correct. You can try to substitute A=30 degrees to (cos^2A - sin^2A / cosA) + (sin^2A - cos^2A / sinA) and (Sin A + cosA), you'll get 2 different values.
     
  6. Jul 31, 2003 #5
    sorry the LHS was this:

    (cos^2A - sin^2A / cosA) + (sin^2A - cos^2A / sinA) + sinA + cosA
     
  7. Jul 31, 2003 #6
    If (cos^2A - sin^2A / cosA) + (sin^2A - cos^2A / sinA) + sinA + cosA = sin A + cos A

    then,
    (cos^2A - sin^2A / cosA) + (sin^2A - cos^2A / sinA) = 0
    which is also not correct

    Your orginal question is to prove
    cosA / (1 - tanA) + sinA / (1 - cotA) = Sin A + cosA

    Let's focus on cosA / (1 - tanA) first. After changing tan A to sin A/cos A, what do you get ? (Hint: you need to simply the denominator first)
     
  8. Jul 31, 2003 #7
    you get:

    (cosA / 1- (sinA/cosA)) + (sinA / 1- (cosA/sinA))

    = cosA ( 1- (cosA/sinA)) + sinA (1- (sinA/cosA))

    = cosA - (cos^2A/sinA) + sinA - (sin^2A/cosA)
     
  9. Jul 31, 2003 #8
    This step is wrong.

    The correct steps should be (I'll only do [cosA/(1 - tanA)] )
    1-tan A
    = 1- (sin A / cos A)
    = (cos A - sin A) / cos A

    therefore,
    cos A / (1- tan A)
    = cos A / [(cos A - sin A) / cos A]
    = cos2A / (cos A - sin A)

    Can you do the second part yourself and simplify the LHS ?
     
  10. Jul 31, 2003 #9
    the second part would be:

    sin^2A / (sinA - cosA)
     
  11. Jul 31, 2003 #10
    Bingo! So can you prove the identity now ? And do you know what you have done wrong?
     
  12. Jul 31, 2003 #11
    yeah, I know what I have done wrong, but how does this prove that identity = sin A + cos A?
     
  13. Jul 31, 2003 #12
    cos2A / (cos A - sin A) + sin2A / (sinA - cosA)
    = cos2A / (cos A - sin A) - sin2A / (cos A - sin A)
    = (cos2A - sin2 A)/(cos A - sin A)
    = (cos A - sin A)(cos A + sin A)/ (cos A - sin A)
    = cos A + sin A
     
  14. Jul 31, 2003 #13
    thanks very much for your help, kl.:wink:
     
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