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Thread: trigonometric identities

  1. #1
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    trigonometric identities

    and i dont understand this at all!

    can you explain this one?

    $\displaystyle \frac{1+sec(-x)}{sin(-x)+tan(-x)}$
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  2. #2
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by singleton2787 View Post
    and i dont understand this at all!

    can you explain this one?

    $\displaystyle \frac{1+sec(-x)}{sin(-x)+tan(-x)}$
    This problems requires that you know about even/odd functions.

    A function is even if $\displaystyle f(-x)=f(x)$, and a function is odd if $\displaystyle f(-x)=-f(x)$.

    Let's apply this to trig:

    $\displaystyle cos(-x)=cos(x)$ implies that cosine is an even function.
    $\displaystyle sin(-x)=-sin(x)$ implies that sine is an odd function.

    Thus, we can rewrite our expression as:
    $\displaystyle \frac{1+sec(-x)}{sin(-x)+tan(-x)}\implies \frac{1+\frac{1}{cos(-x)}}{sin(-x)+\frac{sin(-x)}{cos(-x)}}\implies =\frac{1+\frac{1}{cos(x)}}{-sin(x)-\frac{sin(x)}{cos(x)}}$
    $\displaystyle \implies \frac{1}{-sin(x)}\frac{1+\frac{1}{cos(x)}}{1+\frac{1}{cos(x) }}=-\frac{1}{sin(x)}=-csc(x) \ \forall \ x\neq \frac{k\pi}{2} \ where \ k \ is \ odd$

    Hope this helped you out!
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  3. #3
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    question

    alright thanks that helped a lot
    but i dont understand how you got from here....

    $\displaystyle \frac{1+\frac{1}{cos(x)}}{-sin(x)-\frac{sin(x)}{cos(x)}}$

    to here...

    $\displaystyle \frac{1}{-sin(x)}\frac{1+\frac{1}{cos(x)}}{1+\frac{1}{cos(x) }}$

    how did you get $\displaystyle 1+\frac{1}{cos(x)}$ in the bottom of the second fraction?
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  4. #4
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by singleton2787 View Post
    alright thanks that helped a lot
    but i dont understand how you got from here....

    $\displaystyle \frac{1+\frac{1}{cos(x)}}{-sin(x)-\frac{sin(x)}{cos(x)}}$

    to here...

    $\displaystyle \frac{1}{-sin(x)}\frac{1+\frac{1}{cos(x)}}{1+\frac{1}{cos(x) }}$

    how did you get $\displaystyle 1+\frac{1}{cos(x)}$ in the bottom of the second fraction?
    Pull out a common factor of $\displaystyle -sin(x)$

    Thus, $\displaystyle -sin(x)-\frac{sin(x)}{cos(x)}=-sin(x)\left(1+\frac{1}{cos(x)}\right)$

    $\displaystyle \therefore \frac{1+\frac{1}{cos(x)}}{-sin(x)-\frac{sin(x)}{cos(x)}} \implies \frac{1+\frac{1}{cos(x)}}{-sin(x)\left(1+\frac{1}{cos(x)}\right)}\implies \frac{1}{-sin(x)}\cdot \frac{1+\frac{1}{cos(x)}}{1+\frac{1}{cos(x)}}$
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