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HPlourde

  • 2 years ago

what is (cosx/sinx)/sinxcosx simplified

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  1. abb0t
    • 2 years ago
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    \[\frac{ \cos(x) }{ \sin(x) } \times \frac{ \sin(x)\cos(x) }{ 1 }\]

  2. zepdrix
    • 2 years ago
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    \[\large \frac{\left(\dfrac{\cos x}{\sin x}\right)}{\sin x \cos x}\]Woops I dunno if you're looking at it correctly abbot :O Only the top one is a fraction.

  3. zepdrix
    • 2 years ago
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    \[\large \frac{\left(\dfrac{\cos x}{\sin x}\right)}{\sin x \cos x} \qquad = \qquad \left(\frac{\cos x}{\sin x}\right)\frac{1}{\sin x \cos x}\]

  4. HPlourde
    • 2 years ago
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    ya but what about multiplying by the recipricol.... cause i am seriously confused.

  5. zepdrix
    • 2 years ago
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    \[\large \left(\frac{\cancel{\cos x}}{\sin x}\right)\frac{1}{\sin x \cancel{\cos x}}\]

  6. zepdrix
    • 2 years ago
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    Ok here's the thing with the reciprocal :) lemme explain.

  7. HPlourde
    • 2 years ago
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    ok sounds good

  8. zepdrix
    • 2 years ago
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    The bottom fraction is actually this,\[\large \frac{\sin x \cos x}{1}\] So if you wanted to write it as a division of fractions, you could write it like this,\[\large \frac{\left(\dfrac{\cos x}{\sin x}\right)}{\sin x \cos x} \qquad = \qquad \frac{\left(\dfrac{\cos x}{\sin x}\right)}{\left(\dfrac{\sin x \cos x}{1}\right)}\]And then from here, we could multiply by the reciprocal of the bottom fraction.

  9. HPlourde
    • 2 years ago
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    aha!!!!! now it's making sense!!!

  10. zepdrix
    • 2 years ago
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    When we started this problem, we weren't dividing fractions. Only the top one was a fraction. Maybe that's why there was a little confusion :)

  11. HPlourde
    • 2 years ago
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    i think so...this is the only one that i've been confused over so far... thank you:)

  12. HPlourde
    • 2 years ago
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    i got the right answer. thanks a billion

  13. zepdrix
    • 2 years ago
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    yay!

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