1. Not finding help here? Sign up for a free 30min tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
     

    Attached Files:

    Last edited: Mar 13, 2005
  2. jcsd
  3. Mar 13, 2005 #2
    I can't access the attachment you posted....
     
  4. Mar 13, 2005 #3
    Last edited: Mar 13, 2005
  5. Mar 13, 2005 #4

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    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: 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
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Trig Proofs
  1. A Trig Proof (Replies: 5)

  2. Trig proof (Replies: 2)

  3. Trig proof question (Replies: 2)

  4. Trig Proof (Replies: 5)

Loading...