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Lauren01

  • 3 years ago

What is the exact value of cos((19π)/(12))

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  1. itsmylife
    • 3 years ago
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    0.2588190451 ;)

  2. Lauren01
    • 3 years ago
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    thats wrong..

  3. ZeHanz
    • 3 years ago
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    Try this:\[\cos \frac{ 19 \pi }{ 12 }=\cos \left( \frac{ 16\pi }{ 12 }+\frac{ 3\pi }{ 12 } \right)\] Now simplify the fractions. You wil get numbers we all know and love ;) Then use the formula for cos(a+b).

  4. itsmylife
    • 3 years ago
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    nope that aint wrong ;)

  5. ZeHanz
    • 3 years ago
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    @itsmylife: it's wrong because you left out infinitely many decimals... It is just a (good) approximation.

  6. Lauren01
    • 3 years ago
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    \[\frac{ \sqrt{6}+\sqrt{2} }{ 4 }\]

  7. Lauren01
    • 3 years ago
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    is that correct?

  8. ZeHanz
    • 3 years ago
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    I got \[\frac{ \sqrt{6}-\sqrt{2} }{ 4 }\]

  9. itsmylife
    • 3 years ago
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    well she asked me exact value i used calculator and gave her what she wanted ;)

  10. Lauren01
    • 3 years ago
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    can you explain ? & @itsmylife but thats approximent. exact value is usually always fractions.

  11. itsmylife
    • 3 years ago
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    alright m gonna give ya fraction ;)

  12. itsmylife
    • 3 years ago
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    647/2500 ;) this is way too exact

  13. ZeHanz
    • 3 years ago
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    cos(a+b)=cos(a)cos(b)-sin(a)sin(b) Here we have \[a=\frac{ 16\pi }{ 12 }=\frac{ 4 }{ 3 }\pi\]and \[b=\frac{ 3\pi }{ 12 }=\frac{ 1 }{ 4 }\pi\] Substitute these values:\[\cos \frac{ 4 }{ 3 }\pi \cos \frac{ 1 }{ 4 }\pi-\sin \frac{ 4 }{ 3 }\pi \sin \frac{ 1 }{ 4 }\pi=\]\[-\frac{ 1 }{ 2 }\cdot \frac{ 1 }{ 2 }\sqrt{2}--\frac{ 1 }{ 2 }\sqrt{3}\cdot \frac{ 1 }{ 2 }\sqrt{2}=\frac{ 1 }{ 4 }\sqrt{6}-\frac{ 1 }{ 4 }\sqrt{2}=\frac{ \sqrt{6}-\sqrt{2} }{ 4 }\]

  14. ZeHanz
    • 3 years ago
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    @Lauren01: seems like you missed a little "-"sign somewhere...

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