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Homework Help: Trig Proofs

  1. Mar 13, 2005 #1
    I recently posted about some trig equations, now i'm doing some HW on trig proofs, i got the first couple trig proofs, but had trouble with the last two.
    Here are the two problems (attached). For the first one, i can't even get started. i have some ideas, but i can't find out how to get the right side (1/(1+secθ). For the second one, i simplify the right side for cosθ/sinθ, but i'm not sure if thats right (i can't get the right side to anything like that). any help would be appreciated, thanks!
    http://i2.photobucket.com/albums/y15/seiferseph/untitled.jpg [Broken]

    Attached Files:

    Last edited by a moderator: May 1, 2017
  2. jcsd
  3. Mar 13, 2005 #2
    I can't access the attachment you posted....
  4. Mar 13, 2005 #3
    it says its pending approval, is there anywhere else i can upload it?
    does this work
    http://i2.photobucket.com/albums/y15/seiferseph/untitled.jpg [Broken]
    Last edited by a moderator: May 1, 2017
  5. Mar 13, 2005 #4


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    It's open now- put into latex it says:
    [tex]1. \frac{1-sin^2(\theta)-2cos(\theta)}{cos^2(\theta)-cos(\theta)-2}= \frac{1}{1+ sec(\theta)}[/tex]

    [tex]2. \frac{cos(\theta)-sin(\theta)}{cos(\theta)+sin(\theta)}=sec(2\theta)- tan(2\theta)[/tex]

    Generally speaking, my first step with trig identities is to convert everything into sine and cosine. Here, sec(θ)= 1/cos(θ) so I think that I would in fact convert every thing to cos(θ)- sin(θ) only appears as sin2(θ) in problem 1 so that is just
    [tex]\frac{cos^2(\theta)- 2cos(\theta)}{cos^2(\theta)-cos(\theta)-2}= \frac{1}{1+\frac{1}{cos(\theta)}}= \frac{cos(\theta)}{cos(\theta)+1}[/tex]
    Now think about factoring the terms on the left.
    Last edited by a moderator: Mar 13, 2005
  6. Mar 13, 2005 #5
    thank you, i came close, i had the right side simplified and all (and had the bottom on the left factored), but i couldn't get the top on the left. thanks! now for the 2nd one, is it simplified down to cosθ/sinθ?
  7. Mar 13, 2005 #6
    Neither side simplifiees to [tex]\frac{\cos{\theta}}{\sin{\theta}}[/tex]. You will need to remember what the conjugate of [tex]a-b[/tex] is, if you do it the same way I did. The conjugate of [tex]a-b[/tex] is [tex]a+b[/tex]. Recall that

    [tex]\frac{a+b}{a-b} = \frac{(a+b)^2}{(a-b)(a+b)} = \frac{(a+b)^2}{a^2-b^2}[/tex]

    You might be needing it :)
  8. Mar 13, 2005 #7
    how does the conjugate work? i know you can use it if they are on opposite sides (and diagonal), but how do you use it here. i did multiply the top by (cosθ+sinθ) and then make the bottom (cosθ+sinθ)^2, is that correct? because i solved, but got the reciprocal of the right side.
  9. Mar 13, 2005 #8
    You've made a mistake then. Try multiplying top and bottom of the left side by [tex]\cos{\theta} - \sin{\theta}[/tex] instead, so that the denominator is a difference of squares. You will then need to make several simplifications using identities.

    The way that the conjugate "works" is just multiplication by 1:

    [tex]\frac{a-b}{a+b} = \frac{a-b}{a+b} \cdot \frac{a-b}{a-b} = \frac{(a-b)^2}{a^2-b^2}[/tex]
    Last edited: Mar 13, 2005
  10. Mar 13, 2005 #9
    ok, so you can multiply both top and bottom by either (cosθ + sinθ) or (cosθ - sinθ) to simplify one into the difference of squares?
    thanks! i solved it, getting (1-sin2θ)/(cos2θ) on both sides
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