a maths

Question

solve the following equations for 0 <= x <= 360 ,

sin 3x + cos 3x = sin 2x + cos 2x






the ans is 0, 18, 90, 162, 234, 306 or 360




how to do?! ...thz

Answer 1

利用三角公式
sin A - sin B = 2cos[(A + B)/2]sin[(A - B)/2]
cosA – cosB = - 2sin[(A + B)/2]sin[(A - B)/2]
cos(A+B) = cosAcosB - sinAsinB

1. sin3x+cos3x=sin2x+co s2x
sin3x – sin2x + cos3x – cos2x = 0
2cos[(3x + 2x)/2]sin[(3x – 2x)/2] + 2sin[(3x + 2x)/2]sin[(3x – 2x)/2] = 0
2cos(2.5x)sin(0.5x) - 2sin(2.5x)sin(0.5x) = 0
2[cos(2.5x) – sin(2.5x)]sin(0.5x) = 0 [全式除 cos(2.5x)]
2[1 – tan(2.5x)]sin(0.5x) = 0
sin 0.5x = 0 或 tan(2.5x) = 1
當 sin(0.5x) = 0 則
0.5x = 0 或 0.5x = 180
所以 x = 0o 或 x = 360o
當 tan(2.5x) = 1
2.5x = 45 或 2.5x = 225 或 2.5x = 405 或 2.5x = 585 或 2.5x = 765
x = 18o 或 x = 90o或 x = 162o或 x = 234o或 x = 306o

Answer 2

sin3x-sin2x=cos2x-cos3x
by sum-to-product formulae
sinx-siny=2cos(x+y)/2sin(x-y)/2 and cosx-cosy=-2sin(x+y)/2sin(x-y)/2

so 2cos(5x/2)sin(x/2)=-2sin(5x/2)sin(-x/2)
2cos(5x/2)sin(x/2)=2sin(5x/2)sin(x/2)
1=sin(5x/2)/cos(5x/2)
1=tan(5x/2)
x=18
as 0 <= x <= 360
so x can be 0, 18, 90, 162, 234, 306 or 360

Answer 3

你要用
sin(a)-sin(b)=2 sin[(a+b)/2]．sin[(a-b)/2]
and
cos(a)-cos(b)=-2 sin[(a+b)/2]．sin[(a-b)/2]
Q1
sin 3x + cos 3x = sin 2x + cos 2x

sin 3x - sin 2x- [ cos 2x - cos 3x]=0
by using above formulas, we have
2 cos(2.5x)sin(0.5x) - 2 sin(2.5x)sin(0.5x)=0
2sin(0.5x)[cos(2.5x)-sin(2.5x)]=0
therefore we have
sin(0.5x)=0 or cos(2.5x)=sin(2.5x)
therefore for sin(0.5x)=0： x =0, 360
for cos(2.5x)=sin(2.5x)
tan(2.5x)=1：x=18, 90,162,236,306