數學知識交流 － 解方程

Their is no solution of your question.

對於 sin(x) = cos(y)
可寫作
sin(x) = cos(pi/2 - x) or cos (5pi/2 -x)....
in general,
sin(x) = cos[(4n+1)pi/2 -x] n=int
for one variable x (simplify to one variable)
係E個情況,
tan ( x + (4n+1)pi/2 -x) = tan [(4n+1) pi/2] undefined for n=int

簡單地說就是 sin(x) = cos(y)只會發生在 x+y= 90/270... 的情況,

2012-05-26 01:38:00 補充：
Theorem: 1 = 0 (given that a and b are integers such that a = b + 1)
Proof:
a = b + 1
(a-b)a = (a-b)(b+1)
a2 - ab = ab + a - b2 - b
a2 - ab -a = ab + a -a - b2 - b
a(a - b - 1) = b(a - b - 1)
a = b
b + 1 = b
Therefore, 1 = 0

這題目明顯001的回答是錯的,為甚麼投票是一面倒的?

以下解答單位是 radians,另外n 及k為整數sinx=cosy=tan(x+y)cosy=cos(π/2-x)y=2nπ±(π/2-x)=(2n+1/2)π-x or (2n-1/2)π+xCase (i) tan(x+y)=tan(2n+1/2)π=∞=>無解Case (ii) tan(x+y)=tan[(2n-1/2)π+2x]=-1/tan2x =-cos2x/sin2x =(1-2 cos^2x)/(2 sinx cosx )=sinx=>1-2cos^2x=2cosx sin^2x=2cosx-2 cos^3x=>2cos^3x-2 cos^2x-2 cosx+1=0以三次方程求根公式法或牛頓法,解得三根cosx=1.45160596295578 (不合)cosx=-0.854637679718461=>x=2kπ±2.59564895511614cosx=0.403031716762685=>x=2kπ±1.15596920929611因此方程解為:x=2kπ+2.59564895511614,y=(2n-1/2)π+2.59564895511614x=2kπ-2.59564895511614,y=(2n-1/2)π-2.59564895511614x=2kπ+1.15596920929611,y=(2n-1/2)π+1.15596920929611x=2kπ-1.15596920929611,y=(2n-1/2)π-1.15596920929611