If sinx=3/4, with x in quadrant II, then determine the value of sin 2x

Answer 1

Help I use

S|A
---
T|C

First quad: All trigs. positive
Sec. quad: Only sin and csc
Third quad: Only tan and cot
Fourth quad: Only cos and sec

So, in the second quadrant makes sin positive.

Therefor, I think it should be
sin2x=2*(3/4)

I may be wrong though.

I know if I was looking for the sin(theta)=constant
I would use sin inverse time the constant to find theta the angle.
EX:
sin(theta)=3
theta=sin^(-1)*3

Answer 2

John: sin(theta)=3 cannot be right... the range of sin(theta) is [-1,1].

Lana:
First find the reference angle of x, which is the acute angle formed with the x-axis. In this case, since sin(x)=3/4,
we have
reference angle
=arcsin(3/4)
=48.6 degrees (approx.)

Since we know that x is in the second quadrant,
x=180-reference angle
=180-48.6

2x is therefore twice this amount, or
2(180-48.6)
=360-2*48.6)
=-2*48.6
using the fact that 360-2*48.6 and -2*48.6 are coterminal angles.

sin(2x)=sin(-2*48.6)
=sin(-2*48.6)
=-0.992

Replace 48.6 degrees in the above expressions with the accurate values of arcsin(3/4).

Post if you need more information.