Question

I need help it's pre-calculus!

Answer 1

\[\sin (\frac{ 19\pi }{ 12 })\]

Answer 2

How wuld I put this in degrees to make it easier?

Answer 3

Or do I do something else?

Answer 4

yes

Answer 5

= sin(11pi/6) cos(pi/4) - cos(11pi/6) sin(pi/4)

Answer 6

like that?

Answer 7

yes

Answer 8

pi = 180 degrees,
pi/12=15 degrees

19pi/12 = 285 degrees

285 degrees = 270 + 15 degrees

Using double-angle formula, you will get:
sin(270+15)
= (sin270)(cos15) + (cos270)(sin15)
= (-1)(cos15) + (0)(sin15)
= - (cos15)

15 degrees = 45 - 30 degrees

Thus, by again using the double-angle formula,

-(cos15)
= -(cos45cos30 + sin45sin30)
= -(squareroot(2) / 2) x (squareroot(3) / 2) - (squareroot(2) / 2) x (1/2)
= -(squareroot(6) / 4) - (squareroot(2) / 4)
= -(squareroot(6) - squareroot (2)) / 4

Answer 9

see thats how iwas doing them before but they were written in degrees

Answer 10

Sin 285(degrees) = - 0.966, or
Sin(19π / 12) = - 0.966

Answer 11

Yea CarlosGP I agree with that

Answer 12

yea but thats not one of my answers like loveiskey18 i got
= -(squareroot(6) - squareroot (2)) / 4

Answer 13

sin((19pi)/12) = sin((11pi)/6 - pi/4)
= sin(11pi/6) cos(pi/4) - cos(11pi/6) sin(pi/4)
= [-sin(pi/6)] cos(pi/4) - cos(pi/6) sin(pi/4)
= -sin(pi/6) cos(pi/4) - cos(pi/6) sin(pi/4)
= -(1/2)(1/sqrt[2]) - (sqrt[3]/2)(1/sqrt[2])
.= -1 / (2 sqrt[2]) - sqrt[3] / (2 sqrt[2])
= (-1 - sqrt[3]) / (2 sqrt[2])
= (-sqrt[2] - sqrt[6]) / 4

this is one way of doing it

Answer 14

but either way thanks guys!