A-maths

prove that 1+cos2A+sin4A+cos6A= 4cosAcos2Acos3A.henc e find all angles A between 0度and 180度inclusive which satisfy the equation
cos2A+cos4A+cos6A=-1











(1)1+cos2A+sin4A+cos 6A= 4cosAcos2Acos3A
RHS
=4cosAcos2Acos3A
=4[0.5(cos3A+cosA)]cos3A
=2cos²3A+2cosAcos3A
=2cos²3A+cos4A+cos2A
=cos4A+cos2A+1+cos6A
=LHS

(2)cos2A+cos4A+cos6A =-1 (0°<A<180°)
cos2A+cos4A+cos6A =-1
4cosAcos2Acos3A=0cos Acos2Acos3A=0
cosA=0 or cos2A=0 or cos3A=0
A=90° or 2A=90° or 3A=90°
A=90° or 45° or 30°


2007-06-03 09:31:31 補充：
----------------------------------------------更正：part(1)，1+cos2A+cos4A+cos 6A= 4cosAcos2Acos3A 只要題目有錯，過程沒錯。

2007-06-03 09:36:13 補充：
----------------------------------------------更正：part(1)，題目該是1+cos2A+cos4A+cos 6A= 4cosAcos2Acos3A 只是題目有錯，過程沒錯。

should be 1+cos2A+cos4A+cos6A=4cosAcos2Acos3A ?
wrong typing

we have the formula cosM + cosN = 2 cos [ (M + N) / 2 ] cos [ (M - N) / 2 ] ----- (1)

1+cos2A+cos4A+cos6A
= cos0A+cos2A+cos4A+cos6A
= (cos0A + cos6A) + (cos2A + cos4A)
= 2cos(3A)cos(3A) + 2cos(3A)cosA ------ (by (1) 2 times)
= 2cos(3A) [ cos(3A) + cosA ]
= 2cos(3A) 2cos(2A)cosA ------- ( by (1) )
finished

4cosAcos2Acos3A = 0
cosA = 0 or cos2A = 0 or cos3A = 0

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