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