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Lauren01
 3 years ago
What is the exact value of cos((19π)/(12))
Lauren01
 3 years ago
What is the exact value of cos((19π)/(12))

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ZeHanz
 3 years ago
Best ResponseYou've already chosen the best response.1Try 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).

itsmylife
 3 years ago
Best ResponseYou've already chosen the best response.0nope that aint wrong ;)

ZeHanz
 3 years ago
Best ResponseYou've already chosen the best response.1@itsmylife: it's wrong because you left out infinitely many decimals... It is just a (good) approximation.

Lauren01
 3 years ago
Best ResponseYou've already chosen the best response.0\[\frac{ \sqrt{6}+\sqrt{2} }{ 4 }\]

ZeHanz
 3 years ago
Best ResponseYou've already chosen the best response.1I got \[\frac{ \sqrt{6}\sqrt{2} }{ 4 }\]

itsmylife
 3 years ago
Best ResponseYou've already chosen the best response.0well she asked me exact value i used calculator and gave her what she wanted ;)

Lauren01
 3 years ago
Best ResponseYou've already chosen the best response.0can you explain ? & @itsmylife but thats approximent. exact value is usually always fractions.

itsmylife
 3 years ago
Best ResponseYou've already chosen the best response.0alright m gonna give ya fraction ;)

itsmylife
 3 years ago
Best ResponseYou've already chosen the best response.0647/2500 ;) this is way too exact

ZeHanz
 3 years ago
Best ResponseYou've already chosen the best response.1cos(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:\[\cos \frac{ 4 }{ 3 }\pi \cos \frac{ 1 }{ 4 }\pi\sin \frac{ 4 }{ 3 }\pi \sin \frac{ 1 }{ 4 }\pi=\]\[\frac{ 1 }{ 2 }\cdot \frac{ 1 }{ 2 }\sqrt{2}\frac{ 1 }{ 2 }\sqrt{3}\cdot \frac{ 1 }{ 2 }\sqrt{2}=\frac{ 1 }{ 4 }\sqrt{6}\frac{ 1 }{ 4 }\sqrt{2}=\frac{ \sqrt{6}\sqrt{2} }{ 4 }\]

ZeHanz
 3 years ago
Best ResponseYou've already chosen the best response.1@Lauren01: seems like you missed a little ""sign somewhere...
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