kfsdgte的部落格

標題:

a maths

發問:

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

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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

其他解答:

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|||||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, 90,162,236,3060D7DAC4E6DF60DFA