Show that
sinx (sin2x + sin4x + sin6x) = sin3x sin4x
我係甘計既, 但係去到最尾到就唔知點計:
sin x ( sin 2x + sin 4x + sin 6x )
= sin x [ (sin 2x + sin 6x) + sin 4x ]
= sin x ( 2 sin 4x cos 2x + sin 4x )
= sin 4x sin x ( 2 cos 2x + 1 )
= sin 4x sin x [ 2(1 - sin^2x) +1 ]
= sin 4x sin x( 3 - 4sin^2 x)
....
跟住就唔知點計...可唔可以有人幫我諗下好似我甘計既話要點計落去呀???
F4 A-maths Trigo....急
2008-03-02 3:52 am
回答 (2)
2008-03-02 4:23 am
✔ 最佳答案
Apply triple angle formula (sin 3A = - 4sin^3 A + 3 sin A),...
= sin 4x sin x( 3 - 4sin^2 x)
= sin 4x (3 sin x - 4 sin^3 x)
= sin 4x sin 3x
= RHS
2008-03-01 20:32:03 補充：
Mr. tsangjohn1
I don't think this is a wrong apppraoch.
It is your lack of mathematical knowledge.
2008-03-02 4:24 am
No, wrong approach.
LHS
sinx (sin2x + sin6x + sin4x)
= sinx (2sin4xcos2x + sin4x)
= sinx (sin4x)(2cos2x + 1)
= sin4x (2sinxcos2x + sinx) [Multiply 2cos2x + 1 by sinx]
= sin4x (sin3x - sinx + sinx) [Product-to-sum formula]
= sin3x sin4x = RHS
After you made the sin4x, keep your eye on how can you make sin3x. The only method is by Product-to-sum formula and you know x+2x=3x.
Hope it helps.
LHS
sinx (sin2x + sin6x + sin4x)
= sinx (2sin4xcos2x + sin4x)
= sinx (sin4x)(2cos2x + 1)
= sin4x (2sinxcos2x + sinx) [Multiply 2cos2x + 1 by sinx]
= sin4x (sin3x - sinx + sinx) [Product-to-sum formula]
= sin3x sin4x = RHS
After you made the sin4x, keep your eye on how can you make sin3x. The only method is by Product-to-sum formula and you know x+2x=3x.
Hope it helps.
收錄日期: 2021-04-13 15:13:49
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