If sinx=3/4, with x in quadrant II, then determine the value of sin 2x
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
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.
To determine the value of sin 2x, we can use the double angle formula for sine:
sin(2x) = 2 * sin(x) * cos(x)
First, we need to find the value of cos(x). Since x is in quadrant II, sin(x) is positive while cos(x) is negative. We can use the Pythagorean identity to find cos(x):
cos^2(x) = 1 - sin^2(x)
cos^2(x) = 1 - (3/4)^2
cos^2(x) = 1 - 9/16
cos^2(x) = 16/16 - 9/16
cos^2(x) = 7/16
Taking the square root of both sides:
cos(x) = ±sqrt(7/16)
Since x is in quadrant II, where cos(x) is negative, we take the negative value:
cos(x) = -sqrt(7/16)
cos(x) = -sqrt(7)/4
Now we can substitute the values of sin(x) and cos(x) into the double angle formula:
sin(2x) = 2 * sin(x) * cos(x)
sin(2x) = 2 * (3/4) * (-sqrt(7)/4)
sin(2x) = -6sqrt(7)/16
sin(2x) = -3sqrt(7)/8
Therefore, the value of sin 2x is -3sqrt(7)/8.
To determine the value of sin 2x, we can use the double-angle formula for sine:
sin 2x = 2sin x cos x
Given that sin x = 3/4, we need to find the value of cos x in quadrant II.
In quadrant II, the sine is positive (3/4), and the cosine is negative. To find the cosine value, we can use the Pythagorean identity:
sin^2 x + cos^2 x = 1
Substituting the value of sin x, we get:
(3/4)^2 + cos^2 x = 1
9/16 + cos^2 x = 1
cos^2 x = 16/16 - 9/16
cos^2 x = 7/16
Taking the square root of both sides:
cos x = ± √(7/16)
Since we are in quadrant II, where cosine is negative, we have:
cos x = -√(7/16)
Now we can calculate sin 2x:
sin 2x = 2sin x cos x
= 2 * (3/4) * (-√(7/16))
= -3√7 / 8
Therefore, sin 2x = -3√7 / 8.