F4 A Maths

2007-04-16 12:30 am
試以sin y表示sin 3y
由此,求方程8x^3-6x+1=0的根
答案須準確到2位有效數字

回答 (2)

2007-04-16 12:52 am
✔ 最佳答案
sin3y
= sin(2y+y)
= sin2ycosy + cos2ysiny
= (2sinycosy)cosy + [1 - 2(siny)^2]siny
= 2siny[1 - (siny)^2] + siny - 2(siny)^3
= -4(siny)^3 + 3siny

8x^3 - 6x + 1 = 0
8x^3 - 6x = -1
-8x^3 + 6x = 1

From (a), sub siny into x
-8(siny)^3 + 6siny = 1
2[-4(siny)^3 + 3siny] = 1
2sin3y = 1
sin3y = 1/2
3y = 30° , 150° , 390° , 510° , 750° or 870°
y = 10°, 50°, 130°, 170°, 250° or 290°
siny = sin10°, sin50°, sin130°, sin170°, sin250° or sin290°

Because sin10° = sin170°, sin50° = sin130° and sin250° = sin290°

So, siny = sin10°, sin50° or sin250°

x = siny
= sin10°, sin50° or sin250°
= 0.17, 0.77 or -0.94 (cor. to 2 sig. fig.)
參考: Myself~~~
2007-04-16 12:48 am
sin3y=sin(2y+y)
=sin2ycosy+cos2ysiny
=(2sinycosy)cosy(1-2sin^2 y)siny
=2siny(1-sin^2 y) + siny-2sin^3 y
=3siny-4sin^3 y


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