SOLUTION: Solve sin 2y = cos 4 y for y, where 0 degree <_y<360 degrees

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Question 30135: Solve sin 2y = cos 4 y for y, where 0 degree <_y<360 degrees
Answer by sdmmadam@yahoo.com(530) About Me  (Show Source):
You can put this solution on YOUR website!
sin 2y = cos 4 y for y, where 0 degree <_y<360 degrees
sin 2y = cos 4 y (0 degree <_y<360 degrees )
That is cos(4y) = sin(2y)----(1)
That is cos(4y) = cos[90-2y]
That is cos(4y) = cosk Where k= (90-2y)----(*)
This implies 4y = 2nX(180)+ k or 4y = 2nX(180)- k
And here n =0 as by data (0 degree <_y<360)
(using cosp = cosq implying p=2nX(180)+or-(q) where n is the number of revoultutions round the origin)
That is either 4y = k or 4y = - k
That is either 4y = (90-2y)or 4y = -(90-2y)
That is 4y+2y = 90degrees or 4y = -90+2y
That is 6y = 90 or (4y-2y) =-90
That is y =90/6 = 15 deg or 2y = -90 giving y = -90/2 = -45deg
And (-45) deree = 360-45 = 315 degree
Therefore Answer: y = 15 deg or y = 315 deg
Verification: 1)y = 15 in (1)
LHS = cos(60) = 1/2 = sin(30) = sin(2y) = RHS for y = 15
2)y = 315 in (1)
LHS = cos(4X315) = cos(1260) = cos[3X360+180] = cos(180) = -1
RHS = sin(2X315)=
sin(630) = sin (360+270) = sin(270) = sin(180+90) = -sin(90)= -1 = RHS
Therefore our values 15 deg and 315 deg for y are correct.