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

1. itsmylife

0.2588190451 ;)

2. Lauren01

thats wrong..

3. ZeHanz

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

nope that aint wrong ;)

5. ZeHanz

@itsmylife: it's wrong because you left out infinitely many decimals... It is just a (good) approximation.

6. Lauren01

$\frac{ \sqrt{6}+\sqrt{2} }{ 4 }$

7. Lauren01

is that correct?

8. ZeHanz

I got $\frac{ \sqrt{6}-\sqrt{2} }{ 4 }$

9. itsmylife

well she asked me exact value i used calculator and gave her what she wanted ;)

10. Lauren01

can you explain ? & @itsmylife but thats approximent. exact value is usually always fractions.

11. itsmylife

alright m gonna give ya fraction ;)

12. itsmylife

647/2500 ;) this is way too exact

13. ZeHanz

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

@Lauren01: seems like you missed a little "-"sign somewhere...