Here's the question you clicked on:
Lauren01
What is the exact value of cos((19π)/(12))
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).
nope that aint wrong ;)
@itsmylife: it's wrong because you left out infinitely many decimals... It is just a (good) approximation.
\[\frac{ \sqrt{6}+\sqrt{2} }{ 4 }\]
I got \[\frac{ \sqrt{6}-\sqrt{2} }{ 4 }\]
well she asked me exact value i used calculator and gave her what she wanted ;)
can you explain ? & @itsmylife but thats approximent. exact value is usually always fractions.
alright m gonna give ya fraction ;)
647/2500 ;) this is way too exact
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:\[\cos \frac{ 4 }{ 3 }\pi \cos \frac{ 1 }{ 4 }\pi-\sin \frac{ 4 }{ 3 }\pi \sin \frac{ 1 }{ 4 }\pi=\]\[-\frac{ 1 }{ 2 }\cdot \frac{ 1 }{ 2 }\sqrt{2}--\frac{ 1 }{ 2 }\sqrt{3}\cdot \frac{ 1 }{ 2 }\sqrt{2}=\frac{ 1 }{ 4 }\sqrt{6}-\frac{ 1 }{ 4 }\sqrt{2}=\frac{ \sqrt{6}-\sqrt{2} }{ 4 }\]
@Lauren01: seems like you missed a little "-"sign somewhere...