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Few trig problems

  1. May 11, 2006 #1
    ok here is the first problem:

    sin^2 xcos^4 x and you are to simplify it, the answer is 1/32(2+cos2x-2cos4x-cos6x)

    i will skip a few steps just to save some time:

    1/8(1+2cos2x + cos^2 2x- cos2x -2cos^2 2x -cos^3 2x)

    1/8( 1+2cos2x + 1/2(1+cos4x) - cos2x- 1+cos4x - 1/2cosx(1+cos4x) )

    1/16( 2+ 4cos2x + 1 +cos4x - 2 cos2x - 2 + 2cos4x - cosx(1+cos4x))

    and the rest just gives me a headache, any ideas where to go from there

    the next one:

    (1-secx)/(-sinx - tanx) = -cscx

    prove the identity

    would you multiply by the conjugate or use an identity.. im not sure really how to start it. so far what i have tried it doesnt look like it leads anywhere

    and the final one:

    sqrt((1-cos^2 x)/(1+cos^2 x)) = sqrt(2)|sin x/2|

    prove the identity

    on the right side i have



    im not sure how to simplify the left, i thought about using the conjugate, but it didnt look like it led anywhere

    so if you can help with any of these, please do thanks
    Last edited: May 11, 2006
  2. jcsd
  3. May 12, 2006 #2
    For the first question, you may want to start working on the 2nd expression [tex]\frac{1}{32}(2+cos2x-2cos4x-cos6x) [/tex] first. After all, this question is about simplifying an expression and this definitely looks less simple than the other!

    For the second question, are you supposed to prove [tex] \frac{1-secx}{-sinx-tanx}\ =\ -cosecx [/tex]? If I am not mistaken, this is not an identity. Did you copy the question correctly?

    Same goes for the 3rd question. Are you supposed to prove [tex] \sqrt{\frac{1-cos^{2}x}{1+cos^{2}x}}\ = \sqrt{2}\ \mid sin\frac{x}{2}\mid [/tex]? Why not take another look at the question?
    Last edited: May 12, 2006
  4. May 12, 2006 #3


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    For the first one, you have the right idea. For further simplification, use Factor Formulae.
  5. May 12, 2006 #4
    If I am not mistaken, the second question should look like
    [tex] \frac{1+secx}{-sinx-tanx}\ =\ -cosecx [/tex]
  6. May 12, 2006 #5
    the original for that was (1 + sex(-x))/sin-x) + tan(-x) = -cscx
  7. May 12, 2006 #6
    yes for last two you are to prove the identity

    im not really sure what you mean about the first one? a good part of it my teacher said was right and he said from there on it was jsut an algebra mess
  8. May 12, 2006 #7


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    Can you recheck your second, and your third problem. It does not look right to me.
    I think the second problem should look like:
    [tex]\frac{1 + \sec x}{- \sin x - \tan x} = -csc x[/tex] (as arunbg has already pointed out)
    And the third problem should read:
    [tex]\sqrt{\frac{1 - \cos ^ 2 x}{1 + \cos x}} = \sqrt{2} \left| \sin \left( \frac{x}{2} \right) \right|[/tex]
    Have you given any of the 2 problems above a try?
    I'll give you a hint to start off the second problem:
    You can either multiply boths sides of the equation by the LHS's denominator (i.e -sin x - tan x), or you can try to convert every tan, csc, and sec functions to just sin, and cos functions, and try factoring to get the RHS.
    Give the third problem a try, it won't do you any harm. :)
    Do you know the Double-Angle formulae?
    Can you go from here? :)
  9. May 12, 2006 #8

    the first problem is correctly written the way i had it, but i did mess up it should be cos raised to the first on the bottom

    i have tried these problems forever, a few days at least, we had two assignments 1-71 every other odd and 2-116 every other even... these are the only three i just cant seem to get...
  10. May 12, 2006 #9
    sec(-x)=sec(x) and not -sec(x).
    Third question, Viet Dao's right about the correction.
  11. May 12, 2006 #10


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    Ack, have you looked at the hints I've previously posted, and others' as well? :confused:
    Why don't you give it another try using my hints? It's just for your own sake that we don't give out complete solutions here, at PF. We, however can guide you through the problem, and help you whenever you get stuck. If you'd like, you can skim through the rules here.
    Now, where do you get stuck? :)
  12. May 12, 2006 #11


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  13. May 12, 2006 #12
    i have all but the first and third done...... for the third i have sqrt(cosx-1) but i dont know how to get rid of the sqrt, am i allowed to square it ?
  14. May 12, 2006 #13


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    I would suggest for the first one (if you haven't solved it yet), start afresh and apply Factor Formulae as needed. I did it and it's solvable within five steps.
  15. May 12, 2006 #14
    what is factor formulae?
  16. May 12, 2006 #15


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    I linked it above.
  17. May 13, 2006 #16


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    Uhmm, as I've pointed out before, for #3, you should use the Double Angle formulae, i.e:
    [tex]\cos (x) = \cos ^ 2 \left( \frac{x}{2} \right) - \sin ^ 2 \left( \frac{x}{2} \right) = 1 - 2 \sin ^ 2 \left( \frac{x}{2} \right) = 2 \cos ^ 2 \left( \frac{x}{2} \right) - 1[/tex].
    Can you go from here? :)
    For problem 1, there are many ways to tackle it. Hence, it's very hard to check your answer if you just give us the second half of your work.
    So if you can, just show your complete work, and we may check it for you.
    Or you can try this way.
    Since there are even power of sin, and cos function in this problem, one should consider the Power Reduction Formulae. But first, you can apply the Double Angle Formulae to somplify the expression a bit.
    So I'll start off the problem for you:
    [tex]\sin ^ 2 x \cos ^ 4 x = \sin ^ 2 x \cos ^ 2 x \cos ^ 2 x = \frac{\sin ^ 2 (2x)}{4} \cos ^ 2 x = ...[/tex]
    Can you go from here? :)
    Don't forget to use the Product To Sum Formulae. :)
  18. May 13, 2006 #17


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    For the OP, that's another name for the Factor Formulae.
  19. May 13, 2006 #18
    oh, i have them written down already
  20. May 13, 2006 #19

    i actually cant understand the third problem... i never even knew you could divide the oringinal double angle identities... this is what i have on the left side = to 1-cosx:


    sqrt(1-cosx)... i tryed to put an identity in...but i dont know if i did it right because it didnt look like it led any where
  21. May 13, 2006 #20

    [tex]\sin ^ 2 x \cos ^ 4 x = \sin ^ 2 x \cos ^ 2 x \cos ^ 2 x = \frac{\sin ^ 2 (2x)}{4} \cos ^ 2 x = ...[/tex]

    im not really sure how you got to the last part of it with sin^2 over 4
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