Asked by Alex
                can you prove this equation for me?
sin2x X sec2x = 2sinx
            
        sin2x X sec2x = 2sinx
Answers
                    Answered by
            Reiny
            
    by "proving" an equation , we usually mean to show that it is an identity.
but it is NOT an identity.
(I usually just try it with an arbitrary angle)
sin2x/cos2s = 2sinx
tan 2x = 2sinx
let x= 30°
LS = tan 60 = √3
RS = 2sin30 = 2(1/2) = 1
LS ≠ RS
So it is not an identity, (all we need is one exception)
If you are solving for x, then
2sinx cosx (1/cos2x) = 2sinx
2sinxcosx(1/(2cos^2x - 1) = 2sinx
2sinxcosx = 2sinx(2cos2x - 1)
2sinxcosx - 2sinx(2cos^2x - 1) = 0
2sinx [ cosx - 2cos^2x + 1] = 0
2sinx = 0 or 2cos^2x - cosx - 1) = 0
sinx = 0
x = 0, 180°, 360°
2cos^2x - cosx - 1) = 0
(2cosx + 1)(cosx - 1) = 0
cosx = -1/2 or cosx = 1
cosx = -1/2
x = 120°, 240°
cosx = -1
x = 270°
x = 0, 180, 360, 120, 240, 270 °
I gave the angles in degrees, and in the domain between 0 and 360
I assume you know how to change them to radians if necessary.
    
but it is NOT an identity.
(I usually just try it with an arbitrary angle)
sin2x/cos2s = 2sinx
tan 2x = 2sinx
let x= 30°
LS = tan 60 = √3
RS = 2sin30 = 2(1/2) = 1
LS ≠ RS
So it is not an identity, (all we need is one exception)
If you are solving for x, then
2sinx cosx (1/cos2x) = 2sinx
2sinxcosx(1/(2cos^2x - 1) = 2sinx
2sinxcosx = 2sinx(2cos2x - 1)
2sinxcosx - 2sinx(2cos^2x - 1) = 0
2sinx [ cosx - 2cos^2x + 1] = 0
2sinx = 0 or 2cos^2x - cosx - 1) = 0
sinx = 0
x = 0, 180°, 360°
2cos^2x - cosx - 1) = 0
(2cosx + 1)(cosx - 1) = 0
cosx = -1/2 or cosx = 1
cosx = -1/2
x = 120°, 240°
cosx = -1
x = 270°
x = 0, 180, 360, 120, 240, 270 °
I gave the angles in degrees, and in the domain between 0 and 360
I assume you know how to change them to radians if necessary.
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