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Math Help - Simplifying trigonometric expressions

  1. #1
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    Simplifying trigonometric expressions

    Two questions for now:

    1. Simplify sin \theta/1+cos \theta * 1-sin^2 \theta/cot \theta.

    I got as far as sin^2 \thetacos \theta/1 + cos \theta. Is there any further i can go??

    2. Simplify sin^3 \theta-cos^3 \theta/sin^2 \theta-cos^2 \theta.

    Can i use difference and sum of cubes here to factorise or is this not possible?

    Thanks!
    Steven
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  2. #2
    Moo
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    Hello,

    Please use parenthesis !!

    wHat exactly is in the denominator ?
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  3. #3
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    can you post what have you done so far?
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  4. #4
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    lol sorry i really struggle with the math tags

    Ok for the first one:

    (sin \theta) (1-sin^2 \theta)
    --------------------------
    (1 + cos \theta) (cot \theta)

    So far all i've been able to do is simplify it somehow to:

    (sin^2 \theta * cos \theta)
    ----------------------
    (1 + cos \theta)

    I don't even know if this is right though. We are supposed to be using trigonometric identities but im a bit rusty with the algebra (which i know should be quite easy)
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  5. #5
    Moo
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    Hmmm, for the latex, use [tex]\frac{numerator}{denominator}[/tex] for a fraction

    You can notice that \sin^2 \theta=1-\cos^2 \theta=(1-\cos \theta)(1+\cos \theta)



    Can i use difference and sum of cubes here to factorise or is this not possible?
    Yes you can ! (it's a difference here )
    And factorise the denominator too : a^2-b^2=(a-b)(a+b)
    Last edited by Moo; July 7th 2008 at 12:45 AM. Reason: big typo... i don't know what came in my mind
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  6. #6
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    Thanks!

    But for the first one doesn't that mean that

    sin^2 \theta*cos \theta
    -----------
    1 + cos \theta

    becomes

    (1-cos^2 \theta)(cos \theta)
    -------------
    1 + cos \theta

    ???
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  7. #7
    Moo
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    Yes, it is the idea ! This is due to the identity \cos^2 \theta+\sin^2 \theta=1

    Then it will yield \frac{(1-\cos \theta)(1+ \cos \theta)(\cos \theta)}{1+\cos \theta}

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  8. #8
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    That's much better!

    So the back of my book tells me the answer is

    cos \theta
    ----------
    1 + cos \theta

    Is this the same thing (obviously a more simplified version)? If so, how do they get to the final stage?

    P.S. Thanks for alerting me to the mistake! I was panicking for a moment there
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  9. #9
    Moo
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    Quote Originally Posted by Stevo_Evo_22 View Post
    P.S. Thanks for alerting me to the mistake! I was panicking for a moment there
    Yes, I'm really sorry for that

    cos \theta
    ----------
    1 + cos \theta

    Is this the same thing (obviously a more simplified version)? If so, how do they get to the final stage?
    This is weird

    It's not the same result at all...



    I'll start from scratch, ok ?
    --------------------------------

    \frac{\sin \theta (1-\sin^2 \theta)}{(1+\cos \theta) \cot \theta}

    =\frac{\sin \theta \cos^2 \theta}{(1+\cos \theta) \cdot \frac{\cos \theta}{\sin \theta}}

    =\frac{\sin \theta \cos \theta}{(1+\cos \theta)} \cdot \sin \theta

    =\frac{\sin^2 \theta \cos \theta}{1+\cos \theta} \neq \frac{\cos \theta}{1+\cos \theta}

    Is there a possible mistake in the book ?
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  10. #10
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    It is extremely possible that there is a mistake in the book (it is full of them). So i will take your word for it (particularly because yours makes sense and i got almost the same answer). That help was invaluable to me, I was so close, yet so far Just one more thing, in the end can you cancel the bottom cos \theta with the top one? or is that one the final answer?

    Ok, so for the second one, i am at this point:

    Sin^2 \theta + cos \thetasin \theta + cos^2 \theta
    -------------------
    sin \theta + cos \theta

    This is after I have factorised it, but this par confuses me, especially the top bit. Am i on the right track or way off?
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  11. #11
    Moo
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    Quote Originally Posted by Stevo_Evo_22 View Post
    It is extremely possible that there is a mistake in the book (it is full of them). So i will take your word for it (particularly because yours makes sense and i got almost the same answer). That help was invaluable to me, I was so close, yet so far Just one more thing, in the end can you cancel the bottom cos \theta with the top one? or is that one the final answer?
    How do you want to simplify the cosines ??

    Isn't the final answer (1-\cos \theta) \cos \theta ?

    Well, it's correct you were on the right track Now you just need some practice ^^

    Ok, so for the second one, i am at this point:

    Sin^2 \theta + cos \thetasin \theta + cos^2 \theta
    -------------------
    sin \theta + cos \theta

    This is after I have factorised it, but this par confuses me, especially the top bit. Am i on the right track or way off?
    Right track
    This is the good factorisation !

    Now you can play with it...

    \sin^2 \theta+\cos \theta \sin \theta+\cos^2 \theta=\underbrace{\sin^2 \theta+{\color{red}2} \cos \theta \sin \theta+\cos^2 \theta}_{a^2+2ab+b^2=\dots ?} {\color{red}-\cos \theta \sin \theta}

    and then \frac{\cos \theta \sin \theta}{\cos \theta+\sin \theta}=\frac{1}{\frac{1}{\cos \theta}+\frac{1}{\sin \theta}}=\frac{1}{\sec \theta+\csc \theta}

    I'm not used to csc and sec, so if you know some identities, use them ^^


    ---------------------------------
    Or you can write \cos^2+\sin^2=1 \implies \sin^2 \theta+\cos \theta \sin \theta+\cos^2 \theta=1+\cos \theta \sin \theta


    So you can have at least 2 displays of the final answer
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  12. #12
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    Quote Originally Posted by Moo View Post
    How do you want to simplify the cosines ??

    Isn't the final answer (1-\cos \theta) \cos \theta ?

    Well, it's correct you were on the right track Now you just need some practice ^^
    Yes definitely ( and that is what i meant by simplifying it, thanks for clearing that up for me)






    Quote Originally Posted by ;Moo164767
    Or you can write \cos^2+\sin^2=1 \implies \sin^2 \theta+\cos \theta \sin \theta+\cos^2 \theta=1+\cos \theta \sin \theta


    So you can have at least 2 displays of the final answer
    Yes this makes it much better, i guess i just needed to rearrange it a bit better. So is this the final answer?
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  13. #13
    Moo
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    Quote Originally Posted by Stevo_Evo_22 View Post
    Yes this makes it much better, i guess i just needed to rearrange it a bit better. So is this the final answer?
    These were only the numerator, so after that, simplify with the denominator.

    For the method you quoted, you can simplify cos sin/(cos+sin) like i showed you above
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  14. #14
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    Thanks, the answer given is
    sin \thetacos \theta
    ---------
    sin \theta + cos \theta

    I got the same, except aat the top im left with 1- cos \thetasin \theta

    What happens to the 1 at the top?
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  15. #15
    Moo
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    Hmm, it should be 1+cos(theta)sin(theta)

    Can you show what you have done ?
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