Question

If sinӨ = -1/4 and Ө terminates in the third quadrant, find the exact value of sin2Ө

Answer 1

You need to use the formula

sin(2x) = 2*sin(x)*cos(x)

so you'll need to find the value of cos(theta) first

Answer 2

cos(theta) = 1

Answer 3

how did you get that?

Answer 4

Put it in the calc O_O

Answer 5

put what in the calc? lol

Answer 6

cos(1/4) ?

Answer 7

Oh no and it's -1/4 not 1/4 right?

Answer 8

and that is .999990480

Answer 9

oh yeah, but you don't do cos(-1/4) either

Answer 10

It comes out to the same thing

Answer 11

you need to solve the following for x

sin^2 + cos^2 = 1

(-1/4)^2 + x^2 = 1

1/16 + x^2 = 1

....

...

x = ??

Answer 12

once you get that, you'll know what cos(theta) is

Answer 13

oh my bad >_<

Answer 14

that's fine

Answer 15

x=sqrt of 15/16 = .9682458366
Did I do that right?

Answer 16

better, I would keep the exact form

Answer 17

since we're in Q3, cosine is negative, so cos(theta) = -sqrt(15/16)

sin(2theta) = 2*sin(theta)*cos(theta)

sin(2theta) = 2*(-sqrt(15/16))*(-1/4)

sin(2theta) = 2*(-sqrt(15)/sqrt(16))*(-1/4)

sin(2theta) = 2*(-sqrt(15)/4)*(-1/4)

sin(2theta) = sqrt(15)/8

Answer 18

Thank you c:

Answer 19

np