Question

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

Answer 1

@ganeshie8

Answer 2

@zepdrix

Answer 3

wut 0_o wuts going on here

Answer 4

i need help with this

Answer 5

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

Answer 6

okay yes

Answer 7

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

Answer 8

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

Answer 9

ok

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

Now simplify +_+

Answer 15

i have trouble simplifying that..

Answer 16

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

Answer 17

cosine is the x coordinate

Answer 18

\[-\sqrt{6}-\sqrt{2}/4?\]

Answer 19

is that it?

Answer 20

@zepdrix