anonymous one year ago Find an exact value. sine of nineteen pi divided by twelve. quantity square root of two minus square root of six divided by four. quantity square root of six minus square root of two divided by four. quantity square root of six plus square root of two divided by four. quantity negative square root of six minus square root of two divided by four. @ganeshie8

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1. anonymous

@ganeshie8

2. anonymous

@zepdrix

3. zepdrix

wut 0_o wuts going on here

4. anonymous

i need help with this

5. zepdrix

$\Large\rm \sin\left(\frac{19\pi}{12}\right)=\sin\left(\frac{19\pi/6}{2}\right)$We can start there, ya? And maybe apply half-angle formula? What do you think? :3

6. anonymous

okay yes

7. zepdrix

$\Large\rm \sin\left(\frac{\color{orangered}{\theta}}{2}\right)=\pm\sqrt{\frac{1-\cos\color{orangered}{\theta}}{2}}$

8. zepdrix

So then for our problem,$\Large\rm \sin\left(\frac{\color{orangered}{19\pi/6}}{2}\right)=\pm\sqrt{\frac{1-\cos\color{orangered}{\frac{19\pi}{6}}}{2}}$

9. anonymous

ok

10. zepdrix

What's cosine of 19pi/6? Hmm that angle is too large.. it's larger than 2pi, so we should try to unwind it by subtracting some 2pi's from it

11. zepdrix

$\Large\rm \frac{19\pi}{6}-2\pi\quad=\quad \frac{19\pi}{6}-\frac{12\pi}{6}=\frac{7\pi}{6}$Ya?

12. zepdrix

So our problem becomes,$\Large\rm \pm\sqrt{\frac{1-\cos\color{orangered}{\frac{7\pi}{6}}}{2}}$

13. zepdrix

All we did was spin around the circle once and land in the same spot, to get a better angle to work with.

14. zepdrix

Now simplify +_+

15. anonymous

i have trouble simplifying that..

16. zepdrix

bust out your unit circle or something.. to get the cosine value :U requires some memorization to get those down

17. zepdrix

cosine is the x coordinate

18. anonymous

$-\sqrt{6}-\sqrt{2}/4?$

19. anonymous

is that it?

20. anonymous

@zepdrix