F4 A-maths Trigo

L.H.S=sinx(sin2x+sin4x+sin6x)
=sinx(2sin3xcosx+2sin3xcos3x)
=sin3x(2sinxcosx+2sinxcos3x)
=sin3x(sin2x+sin4x-sin2x)
=sin3xsin4x
=R.H.S

use Product to Sum Formula
sinxsin2x=1/2(cosx-cos3x)
sinxsin4x=1/2(cos3x-cos5x)
sinxsin6x=1/2(cos5x-cos7x)

sinx (sin2x + sin4x + sin6x)
=sinxsain2x+sinxsin4x+sinxsin6x
=1/2(cosx-cos7x)
=-sin(-3x)sin4x sum to product formula
=sin3xsin4x

sinx (sin2x + sin4x + sin6x)
=sinx (2sin3x cosx+sin6x)
=sinx(2sin3x cosx+2sin3x cos3x)
=2sinx sin3x(cosx+cos3x)
=2sinx sin3x(2cos2x cosx)
=4sinx cosx cos2x sin3x
=2sin2x cos2x sin3x
=sin3x sin4x

2008-03-01 20:47:26 補充：
= sin4x sin x( 3 - 4sin^2 x)
= sin4x (3sinx-4sin^3x)
= sin3x sin4x
(by triple angle formula, now out syl)

不滿的可以按下面這樣做:

2008-03-01 20:47:37 補充：
= sin x( 3 - 4sin^2 x)
= (3sinx-4sin^3 x)
= 2sinx - 2sin^2x + sinx -2sin^2 x
= 2sinx(1-sin^2x) +sinx -2sin^2 x
= 2sinxcos^2x +sinx(1-2sin^x)
= sin2xcosx + sinxcos2x
= sin(x+2x)
= sin3x

if it is reversed, it's the proof for triple angle formula:
sin3x=3sinx-4sin^3x

2008-03-01 20:57:44 補充：
我建議你開頭就eliminate左個sinx, 因為個答案無sinx
另一方面要保留sin4x係括號出面