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Homework Help: Trig identities help! How to simplify by multiplication/division

  1. Dec 7, 2009 #1
    Ok guys, here's my problem. I left on vacation with my parents before I learned how to do these correctly. I have been trying and I sorta have the gist of them down. For instance, tan A*sec A simplified is sin A. A=theta. But as I move on they stop making sense, and this is where my problem occurs .I don't know the correct way to do these, but I tried.

    1) (tan^2 A)( 1 ) + 1 where A again equals theta
    -------- ------
    sec^2 A sin

    2)1+cos A + sin A
    -------- --------
    sin A 1+cos A

    3) sin A * cot A * csc A
    tan A

    Ok, so on the first one my understanding is to first change them into just sin/cos. What i did was:

    (sin A^2) 1 1
    --------- * --------- + ------
    (cos A^2) 1 sin A
    cos A

    from here i think i need to do mult. inv. of the middle to get it to just cos A. This is where i get lost. I want to multiply the first 2 now but I get a huge answer then adding the last term just makes it look horrible.

    On the 2nd one, I cross multiplied and I got

    (1+cos A)^2 + sin
    ------------------ = 1
    (1+cos A)^2 + sin

    On the last one I wrote it out again in cos/sin and got

    sin A cos A 1
    ---------- * --------- * -------
    sin A sin A sin A
    cos A

    Trying to solve this I got

    cos^2 A
    sin^2 A

    If someone could please help me understand these and maybe give me a few pointers I'd very much appreciate it. Thanks!
    1. The problem statement, all variables and given/known data

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Dec 7, 2009 #2


    Staff: Mentor

    No, that's not right. tanA*secA = (sinA/cosA)*(1/cosA) = sinA/(cosA)^2.

    The bottom in the 2nd factor should by 1/(cosA)^2.
    You can't cross multiply - that applies to equations, and what you have is not an equation. You can multiply each expression by 1 to get a common denominator. You have the right numerator, but the denominator should be (1 + cosA)sinA
    You're not supposed to "solve" these - just simplify them. What you ended with is correct, but can be simplified more. Remember that cosA/sinA = cotA.
  4. Dec 7, 2009 #3
    Do you have a right side to any of these equations? Or do you just have to get the equation to the simplest terms?
  5. Dec 7, 2009 #4


    Staff: Mentor

    None of them is an equation, at least as posted here. I think the goal here is to simplify the given expressions.
  6. Dec 8, 2009 #5
    Sorry, I guess that the post messed up what the problem looked like. I'm supposed to simplify them, but I really don't get what to do with tan^2 and how to change everything so that I can simplify. This one is what I mean:

    (tan^2 A)(1/sec^2 A) + 1/sin A

    I think tan^A is sin^2 A/cos^2 A

    But once I hit sec^2 A I get confused. wouldn't the middle end up being just cos^2 A?

    Then from there I don't know how to simplify it because the farthest I get (don't know for sure if its even right) is this:

    sin^2 A/cos^2 A*cos^2 A + 1/sin A
    Last edited: Dec 8, 2009
  7. Dec 8, 2009 #6


    Staff: Mentor

    I can't tell what the problem is. In post 1, you have this in post 5.
    [tex]\frac{tan^2A}{sec^2A} + \frac{1}{sinA}[/tex]

    In post 5 you have this:
    [tex]=~\frac{tan^2A}{\frac{1}{sec^2A}} + \frac{1}{sinA}[/tex]
  8. Dec 8, 2009 #7
    I'm sorry for the confusion, its the first one of those 2. I just don't understand how to simplify it because I don't know the subject very well
  9. Dec 8, 2009 #8


    Staff: Mentor

    [tex]\frac{tan^2A}{sec^2A} + \frac{1}{sinA}[/tex]
    [tex]=\frac{\frac{sin^2A}{cos^2A}}{\frac{1}{cos^2A}} + \frac{1}{sinA}[/tex]
    Can you continue from here?
  10. Dec 8, 2009 #9
    I think so. So it would be:

    (sin^2 A)(cos^2 A)/cos^2 A + 1/sin A then cancel out the cos terms correct?

    so sin^2 A/sin A would be the next step, which I think simplifies to just sin A.

    Is this right?
  11. Dec 8, 2009 #10


    Staff: Mentor

    How are you getting that? You have sin^2(A) + 1/sinA. That's different from sin^2(A) *(1/sinA)
  12. Dec 9, 2009 #11
    Ah, yes your right, I wasn't paying attention to that. Sorry, I don't mean to waste your time, I'll try to get this over with. i think I might have it this time.

    sin^2(A) + 1/sinA is adding so I have to get a common denominator. So, I take the denom. of each term and multiply each by the other term so that they will have the same base.

    so i did sin^2(A)/1 + 1/sinA which after getting a common denominator became

    sin^3(A)/sinA then it simplifies to


    Is this correct?
  13. Dec 9, 2009 #12


    Staff: Mentor

    Up to sin^2(A)/1 + 1/sinA you were right on the money, but then went rapidly downhill. The step you skipped (or at least didn't show) was
    [tex]\frac{sin^3 A}{sin A} + \frac{1}{sin A}[/tex]

    Now, both denominators are the same, so what do you get when you add the two expressions?
  14. Dec 9, 2009 #13
    Oh, I see, so its sin^3(A) + 1/sinA?

    Then it ends up as sin^2(A) + 1

    Is that correct now?
  15. Dec 9, 2009 #14


    Staff: Mentor

    No again. It is (sin^3(A) + 1)/sinA. Now maybe your intent was that sin^(A) + 1 was the numerator and sinA was the denominator, but without parentheses, that's not how it would be interpreted.
    Also no.

    Your problems indicate a lack of understanding of how fractions are added. What we're doing here is no different from the ordinary fraction arithmetic.

    Your first mistake in this post is like saying that 8/2 + 1/2 = 8 + 1/2 instead of (8 + 1)/2.
    8 + 1/2 is 8.5, while (8 + 1)/2 = 9/2 = 4.5.

    Your second mistake (assuming that you meant (sin^3(A) + 1)/sinA) is like saying that (8 + 1)/2 = 4 + 1. I think you might be "cancelling" when you have no justification for doing this. Cancelling is removing identical factors from the numerator and denominator. Factors are expressions that are multiplied together. In this case, neither sinA nor 2 is a factor.

    If I were you, I would definitely brush up on fractions and rational expressions. The work you are doing in precalculus requires that you have solid working skills in these areas.
  16. Dec 9, 2009 #15
    Ya, I see my mistake and understand how to correct it now. I had forgotten that you cant just cancel them out because like you said, I can't justify it. I have been working on these problems for the past 3 days now and I'm finally getting the hang of it. Thank you soooooo much for your help. I've been using what you posted as a guideline and It has helped a lot.
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