# Math

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

1. 👍 0
2. 👎 0
3. 👁 93
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

1. 👍 0
2. 👎 0
posted by John
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).

1. 👍 0
2. 👎 0
posted by MathMate

## Similar Questions

1. ### Trig Help

Prove the following: [1+sinx]/[1+cscx]=tanx/secx =[1+sinx]/[1+1/sinx] =[1+sinx]/[(sinx+1)/sinx] =[1+sinx]*[sinx/(sinx+1)] =[sinx+sin^2x]/[sinx+1] =[sinx+(1-cos^2x)]/[sinx+1] =??? This is where I'm stuck. Can someone help me.

asked by Anonymous on March 11, 2012
2. ### math

suppose sinx=1/5, cosy=2/3, and x and y are in the first quadrant. determine sin (x+y)

asked by sh on March 5, 2009
3. ### Trig

Find sin(x+y), cos(x-y), tan(x+y), and the quadrant of (x+y) if sinx= -1/4, cosy= -4/5, with x and y in quadrant 3.

asked by Brendan on April 10, 2016
4. ### pre calc

Find the exact value of sin(x-y) if sinx=-3/5 in Quadrant III and cosy=5/13 in Quadrant I.

asked by Jennifer on May 21, 2013
5. ### Mathematics - Trigonometric Identities - Reiny

Mathematics - Trigonometric Identities - Reiny, Friday, November 9, 2007 at 10:30pm (sinx - 1 -cos^2x) (sinx + 1 - cos^2x) should have been (sinx - 1 + cos^2x) (sinx + 1 - cos^2x) and then the next line should be sin^2x + sinx -

asked by Anonymous on November 10, 2007
6. ### Math Help

Hello! Can someone please check and see if I did this right? Thanks! :) Directions: Find the exact solutions of the equation in the interval [0,2pi] cos2x+sinx=0 My answer: cos2x+sinx=cos^2x-sin^2x+sinx =1-sin^2x-sin^2x+sinx

asked by Maggie on February 2, 2015
7. ### calculus

Could someone check my reasoning? thanx Find the derivative of the function. sin(sin[sinx]) I need to use the chain rule to solve. So I take the derivative sin(sin[sinx) first. Then multiply that by the inside which is the

asked by XCS on May 7, 2009
8. ### Math

Find sin (x-y) if sinx = -2/5 and siny = 2/3 X is in quadrant IV, and y is in quadrant I

asked by Anonymous on November 6, 2012
9. ### Math (trigonometric identities)

I was given 21 questions for homework and I can't get the last few no matter how hard and how many times I try. 17. Sinx-1/sinx+1 = -cos^2x/(sinx+1)^2 18. Sin^4x + 2sin^2xcos^2x + cos^4x = 1 19. 4/cos^2x - 5 = 4tan^2x - 1 20. Cosx

asked by Kim on December 4, 2012
10. ### Trig

If secx = 8 and -pi/2 < x < 0, find the exact value of sin2x Use the identity sin 2x = 2(sinx)(cosx) if secx = 8, then cosx = 1/8 where x is in the fourth quadrant. consider a right angled triangle with x=1, r=8, then y=?? by

asked by Ashley on April 14, 2007

More Similar Questions