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SugarRainbow

  • 2 years ago

help with secx/sinx-sinx/cosx!!!!!!!!!!!!!!!!!!!!!!!!!!!!

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  1. Shadowys
    • 2 years ago
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    you wanna simplify it? note that sec x = 1/cos x so it becomes \(\frac{1}{sin x cos x} - \frac{sin x}{cos x} \) if you continue to put both fractions into one , you'll notice something beautiful.

  2. SugarRainbow
    • 2 years ago
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    well i got ((1/cosx)/sinx)-sinx/cos x and IDK what what to do after that unless i change cosx to 1/secx then can i cancel out the two secx?

  3. Shadowys
    • 2 years ago
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    The sec x cannot be "cancelled". from above, \[\frac{ 1 }{ \sin x \cos x }-\frac{ \sin x }{ \cos x }=\frac{ 1-\sin x(\sin x) }{ \sin x \cos x}\] and \(cos^2 x+sin^2 x =1 \) :) I think this will help.

  4. SugarRainbow
    • 2 years ago
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    but if you subtract sin from 1 how did you get 1-sinx(sinx)? wouldn't it just be 1-sinx?

  5. Shadowys
    • 2 years ago
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    hmm...I didn't subtract sin x.. \[\frac{ 1 }{ a }-\frac{ 1 }{ b }=\frac{ b-a }{ ab }\] I think this should clear things up.

  6. SugarRainbow
    • 2 years ago
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    but you have two things in the first one cuz it's1/sinxcosx not just 1/sinx

  7. Shadowys
    • 2 years ago
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    yup! :) so I multiplied sin x to \( \frac {sin x}{cos x}\) to have the same denominator.

  8. SugarRainbow
    • 2 years ago
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    oh okay then you seperated it into 1-sinx(sinx) because when you multiply it you get 1-sin^2x

  9. SugarRainbow
    • 2 years ago
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    but now what? cuz i know it has to equal cotx but i don't see how

  10. Shadowys
    • 2 years ago
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    well, since from \(cos^2x+sin^2x=1\) we know that \(cos^2 x=1-sin^2 x\) so it simplifies to \(\frac{cos^2}{sin x cos x}\)=cot x

  11. SugarRainbow
    • 2 years ago
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    but there's a cosx in the denominator so how does that work if cot x =cosx/sinx?

  12. Shadowys
    • 2 years ago
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    well, \( \frac{cos^2 x}{ sinxcosx}\)= \(\frac{cos x cosx}{sin x cos x}\) and we can cancel the cos x from the numerator and denominator.

  13. SugarRainbow
    • 2 years ago
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    oh okay!!!! weLl thanx

  14. Shadowys
    • 2 years ago
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    You're welcome :)

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