Question

sinx=1/3 and x lies in quad. 2, find exact value of sin(x+pi/6)

THANKS!

Answer 1

\(\bf sin(x) = \cfrac{1}{3} \implies \cfrac{\textit{opposite}}{\textit{hypotenuse}} \implies\cfrac{b}{c}\\
c^2 = a^2 + b^2 \implies \sqrt{c^2-b^2} = a\\
\textit{keep in mind that}\\
\color{blue}{sin(\alpha-\beta) = sin(\alpha)cos(\beta)-cos(\alpha)sin(\beta)}\)

Answer 2

also recall that is "x" is in the 2nd Quadrant, "x" is negative, "y" is positive
so the sine will be positive, as you can see it's
but the cosine will be negative

Answer 3

well, I'd say angle "x" is in the 2nd Quadrant and cosine there is negative

Answer 4

lemme word it better :)

also recall that is "GIVEN ANGLE" is in the 2nd Quadrant, "x" is negative, "y" is positive there
so the sine will be positive, as you can see it's
but the cosine will be negative

Answer 5

Thanks but I'm still a bit confused as to how to find x

Answer 6

well, you're not required to find the angle "x"
you're just required to find the exact value of the sine function of the angle "x + pi/6"

Answer 7

and you'd do it by using the coordinate values for "x"

Answer 8

if we know that \(\bf sin(x) = \cfrac{1}{3} \implies \cfrac{\textit{opposite}}{\textit{hypotenuse}} \implies\cfrac{b}{c}\\
c^2 = a^2 + b^2 \implies \sqrt{c^2-b^2} = a\)

what value would you get for "a"?

Answer 9

A= radical 8

Answer 10

\(\bf a = \pm \sqrt{8} \implies a = \pm 2\sqrt{2}\)
keep in mind, "ANGLE x" is in the 2nd Quadrant, so it's cosine will be negative, thus
\(\bf a = -2\sqrt{2}\)

Answer 11

so now that we know that, what would be the cosine of "x"
what's is cos(x)?

Answer 12

.999?

Answer 13

how did you get that?

Answer 14

I put cos(-2rad2) into calculator

Answer 15

hmmm, well, we didn't find the cos(x) btw
we only found the side "a" for the angle "x"

Answer 16

cosine of the angle will be \(\bf cos(x) = \cfrac{\textit{adjacent side}}{\textit{hypotenuse}}\implies\cfrac{a}{c}\)

Answer 17

Ohhh ok

Answer 18

so, what's the hypotenuse for "x"?

Answer 19

Sry im notat home right now soim responding from my phone

Answer 20

ok

Answer 21

wow thanks so much! i finally got it and got the right answer :D