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Thread: Simplifying trigonometric expressions

  1. #1
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    Simplifying trigonometric expressions

    Two questions for now:

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

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

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

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

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

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

    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 $\displaystyle \sin^2 \theta=1-\cos^2 \theta=(1-\cos \theta)(1+\cos \theta)$



    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 : $\displaystyle a^2-b^2=(a-b)(a+b)$
    Last edited by Moo; Jul 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$\displaystyle \theta$*cos$\displaystyle \theta$
    -----------
    1 + cos$\displaystyle \theta$

    becomes

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

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

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

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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$\displaystyle \theta$
    ----------
    1 + cos$\displaystyle \theta$

    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$\displaystyle \theta$
    ----------
    1 + cos$\displaystyle \theta$

    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 ?
    --------------------------------

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

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

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

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

    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$\displaystyle \theta$ with the top one? or is that one the final answer?

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

    Sin^2$\displaystyle \theta$ + cos$\displaystyle \theta$sin$\displaystyle \theta$ + cos^2$\displaystyle \theta$
    -------------------
    sin$\displaystyle \theta$ + cos$\displaystyle \theta$

    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$\displaystyle \theta$ with the top one? or is that one the final answer?
    How do you want to simplify the cosines ??

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

    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$\displaystyle \theta$ + cos$\displaystyle \theta$sin$\displaystyle \theta$ + cos^2$\displaystyle \theta$
    -------------------
    sin$\displaystyle \theta$ + cos$\displaystyle \theta$

    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...

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

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

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


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


    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 $\displaystyle (1-\cos \theta) \cos \theta$ ?

    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 $\displaystyle \cos^2+\sin^2=1 \implies \sin^2 \theta+\cos \theta \sin \theta+\cos^2 \theta=1+\cos \theta \sin \theta$


    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$\displaystyle \theta$cos$\displaystyle \theta$
    ---------
    sin$\displaystyle \theta$ + cos$\displaystyle \theta$

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

    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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