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Math Help - Help me with trig functions!!

  1. #1
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    Angry Help me with trig functions!!

    I'm having a hard time understanding how to verify trig functions using trig identities.

    can someone walk me through how to verify this equation? thanks

    (cscθ)(tanθ) = secθ
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  2. #2
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    Quote Originally Posted by TacticalPro View Post
    I'm having a hard time understanding how to verify trig functions using trig identities.

    can someone walk me through how to verify this equation? thanks

    (cscθ)(tanθ) = secθ
    basic identities which you should know ...

    \csc{x} = \dfrac{1}{\sin{x}}

    \tan{x} = \dfrac{\sin{x}}{\cos{x}}

    \sec{x} = \dfrac{1}{\cos{x}}
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  3. #3
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    It is like this
    (1/sinx)(sinx/cosx)= sec
    (1/cosx)= secx
    secx=secx
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  4. #4
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    I can help

    \displaystyle \csc \theta \tan \theta

    \displaystyle = \frac{1}{\sin \theta}\times \frac{\sin \theta}{\cos \theta}

    \displaystyle =\frac{1}{\cos \theta}

    \displaystyle =\sec\theta
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  5. #5
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    okay thanks! what about this problem?

    csc[(pi/2)-θ] / tan(-θ) = -cscθ

    i reduced that down to

    secθ / tanθ = cscθ

    and i'm stuck. can anyone help me with what I do next? thanks
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  6. #6
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    Quote Originally Posted by TacticalPro View Post
    okay thanks! what about this problem?

    csc[(pi/2)-θ] / tan(-θ) = -cscθ

    i reduced that down to

    secθ / tanθ = cscθ

    and i'm stuck. can anyone help me with what I do next? thanks
    No,

    you should have

    \displaystyle\frac{sec\theta}{tan(-\theta)}=csc\theta}

    Now write tan in terms of sine and cosine and use sin(-A)=-sinA, cos(-A)=cosA.
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  7. #7
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    Quote Originally Posted by Archie Meade View Post
    No,

    you should have

    \displaystyle\frac{sec\theta}{tan(-\theta)}=csc\theta}

    Now write tan in terms of sine and cosine and use sin(-A)=-sinA, cos(-A)=cosA.
    I'm assuming you forgot to put a negative sign (-) in front of cscθ?

    anyway, i did what u suggested and I got:

    secθ / (-sinθ/cosθ) = - cscθ

    and I rewrote it as

    secθcosθ / -sinθ = - cscθ

    and now I'm stuck again
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  8. #8
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    No, I didn't forget.
    The minus sign only comes at the last step.
    I'm working from the left and evaluating that.

    Now, what is sec\theta cos\theta\;\;?
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  9. #9
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    Quote Originally Posted by Archie Meade View Post
    No, I didn't forget.
    The minus sign only comes at the last step.
    I'm working from the left and evaluating that.

    Now, what is sec\theta cos\theta\;\;?
    Well, isn't it 1? because:

    cosθ/cosθ
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  10. #10
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    Quote Originally Posted by TacticalPro View Post
    Well, isn't it 1? because:

    cosθ/cosθ
    Correct.
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  11. #11
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    ok so now i'm left with:

    1/-sinθ = -cscθ

    cscθ = cscθ
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  12. #12
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    Quote Originally Posted by TacticalPro View Post
    ok so now i'm left with:

    1/-sinθ = -cscθ

    cscθ = cscθ
    Why not just say

    \displaystyle\frac{1}{-\sin}=-\csc=\text{RHS}
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  13. #13
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    You're probably overanalysing the two sides being equal.

    You only need evaluate the left hand side to see what it eventually yields...

    \displaystyle\frac{1}{-sin\theta}=-\frac{1}{sin\theta}=-csc\theta
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