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Trig Identity Proof Check

  1. Jan 5, 2014 #1
    Hi there,

    This is my very first post, so I'd like to say thanks for reading and hi basically. :biggrin:

    I'm relatively confident my attempt at the proof is correct, but since the method is quite different from other examples I have seen, it kind of makes me nervous. I was hoping someone could verify my answer?

    1. The problem statement, all variables and given/known data

    Prove that:

    [itex] \sin^2\theta (1 + sec^2\theta) = sec^2 \theta - cos^2 \theta[/itex]

    2. Relevant equations


    3. The attempt at a solution

    So I try manipulate the right hand side to take exactly the same form as the left. First I rewrite the terms as their 'ordinary' equivalents:

    ##\dfrac{1}{\cos^2 \theta} - \cos^2 \theta##

    ##\dfrac{1}{\cos^2 \theta} + \sin^2 \theta - 1##

    I then combine the trig terms:

    ##\dfrac{\sin^2 \theta \cos^2 \theta + 1}{\cos^2 \theta} - 1##

    Now I assimilate the constant term of -1:

    ##\dfrac{\sin^2 \theta \cos^2 \theta - \cos^2\theta + 1}{\cos^2 \theta}##

    I now use the fact that ##-\cos^2 \theta + 1 = \sin^2 \theta##:

    ##\dfrac{\sin^2 \theta \cos^2 \theta + \sin^2 \theta}{\cos^2 \theta}##

    I remove the common factor of ##\sin^2 \theta## from the numerator:

    ##\dfrac{\sin^2 \theta( \cos^2 \theta + 1)}{\cos^2 \theta}##

    Next I split the fraction into its constituent products:

    ##\sin^2 \theta \left(\dfrac{\cos^2 \theta + 1}{\cos^2 \theta} \right)##

    Simplifying:

    ##\sin^2 \theta \left(1 + \dfrac{1}{\cos^2 \theta} \right) \equiv \sin^2\theta (1 + sec^2\theta)##

    Have I gone wrong anywhere at all or does this work just fine? Thanks a lot!

    ~Snooz
     
  2. jcsd
  3. Jan 5, 2014 #2
    No, you do not. You did not remove anything at all. If you had done so, it would be wrong.
    I may be wrong about this.

    Works just fine.
    I may be wrong about this, too.

    Doubtful poster is doubtful.
     
  4. Jan 5, 2014 #3
    It looks good.

    If i were to rewrite the LHS as ##\sin^2\theta+\tan^2\theta##, do you (a) see how I got that and (b) see how to get to the RHS in two very straightforward steps?

    Sometimes with these types of problems, it's "fun" to find alternate ways of doing the problems that are potentially easier. It's a skill that'll come in handy down the road if you continue on to calculus.
     
  5. Jan 5, 2014 #4
    Thanks for replying! Yeah it struck me that when I wrote 'removed' that maybe didn't sound exactly right. What I mean to say is I factored the numerator using that common factor of ##\sin^2 \theta##.

    Thanks, I'm glad it looks all right. I was conscious when I was doing it this was a very long way round of doing it, and I'm very new to using reciprocal trigonometric functions so it isn't quite free flowing yet. In fact, trigonometry is one area of mathematics I haven't given equal attention, and for no good reason.

    Haha! Yes, for (a) you just expanded the bracket, and I see how you arrived at (b):

    ##\sin^2 \theta + \dfrac{\sin^2 \theta}{\cos^2 \theta}##

    Since ##\tan^2 \theta = \sec^2 \theta - 1##:

    ##\sin^2 \theta + \dfrac{1}{\cos^2 \theta} - 1##

    Also ##\sin^2 \theta - 1 = - \cos^2 \theta##, so:

    ##\dfrac{1}{\cos^2 \theta} - \cos^2 \theta##

    Geez! That was much quicker. I'm going to need to put a good bit more practice in here. Thank you very much guys.
     
  6. Jan 5, 2014 #5

    AlephZero

    User Avatar
    Science Advisor
    Homework Helper

    The left hand side contains ##\sin## and ##\sec##, and the right hand size has ##\cos## and ##\sec##.

    So just shoot to kill and change the ##\sin## to ##\cos##.

    LHS = ##(1 - \cos^2\theta)(1 + \sec^2\theta)## and multiply out :smile:
     
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