How do you solve sin x = 2/5 without a graph?
vanush kdslkf Joined Oct 10, 2004 Messages 546 Location Sydney, Australia Gender Male HSC 2006 Jul 4, 2006 #1 How do you solve sin x = 2/5 without a graph?
dddman Banned Joined Jun 13, 2006 Messages 120 Gender Male HSC 2005 Jul 4, 2006 #2 x = sin^-1(2/5) = 23.57Deg or .13pi Rads
vanush kdslkf Joined Oct 10, 2004 Messages 546 Location Sydney, Australia Gender Male HSC 2006 Jul 4, 2006 #3 sine inverse 2/5 gives .13pi c.. but how do you find the other point (because the question asks for values between 0 and 2pi) Last edited: Jul 4, 2006
sine inverse 2/5 gives .13pi c.. but how do you find the other point (because the question asks for values between 0 and 2pi)
P pLuvia Guest Jul 4, 2006 #4 Since sinx is positive in the first and second quadrant then sinx=2/5 x=sin-1(2/5) and 180-sin-1(2/5) =23.58 and 156.42
Since sinx is positive in the first and second quadrant then sinx=2/5 x=sin-1(2/5) and 180-sin-1(2/5) =23.58 and 156.42
vanush kdslkf Joined Oct 10, 2004 Messages 546 Location Sydney, Australia Gender Male HSC 2006 Jul 4, 2006 #5 o yeah lol
vanush kdslkf Joined Oct 10, 2004 Messages 546 Location Sydney, Australia Gender Male HSC 2006 Jul 5, 2006 #6 Show, with a diagram, that cos (3pi/2 + theta) = sin theta
alcalder Just ask for help Joined Jun 26, 2006 Messages 597 Location Sydney Gender Female HSC N/A Jul 5, 2006 #7 Graph y = sin (theta) y = cos(3pi/2 + theta) = cos(theta - (-3pi/2)) The second curve is just cos(theta) shifted 3pi/2 to the left. You will see that the graphs for both equations are the same. Is that what you were wanting??
Graph y = sin (theta) y = cos(3pi/2 + theta) = cos(theta - (-3pi/2)) The second curve is just cos(theta) shifted 3pi/2 to the left. You will see that the graphs for both equations are the same. Is that what you were wanting??
vanush kdslkf Joined Oct 10, 2004 Messages 546 Location Sydney, Australia Gender Male HSC 2006 Jul 6, 2006 #8 yes, but i was thinking you had to show it using the astc diagram because the exercise it is in is before graph drawing Last edited: Jul 6, 2006
yes, but i was thinking you had to show it using the astc diagram because the exercise it is in is before graph drawing
R Riviet . Joined Oct 11, 2005 Messages 5,584 Gender Undisclosed HSC N/A Jul 6, 2006 #9 Using the diagram, cos(3pi/2 + θ)=cos[2pi - (pi/2 - θ)] =cos(pi/2 - θ) since cos is positive in first and fourth quadrants, ie cos(2pi-A)=cosA =sinθ
Using the diagram, cos(3pi/2 + θ)=cos[2pi - (pi/2 - θ)] =cos(pi/2 - θ) since cos is positive in first and fourth quadrants, ie cos(2pi-A)=cosA =sinθ