yosi asked in 科學數學 · 1 decade ago

求sin(π/7)*sin(2π/7)*sin(3π/7)

證明sin(π/7)*sin(2π/7)*sin(3π/7)=√(7)/8

不要用到複數和隸美福定理

盡量用三角函數的公式去算

給最快的 謝謝

4 Answers

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  • 1 decade ago
    Favorite Answer

    You are looking for P = sin(π/7)sin(2π/7)sin(3π/7)

    Let s = sin(π/7), c = cos(π/7) so s + c = 1.

    From cos(α + β) and sin(α + β) formulas come

    cos(2π/7) = c - s = 2c - 1

    sin(2π/7) = 2cs

    sin(3π/7) = s(2c - 1) + c(2cs) = s(4c - 1), and

    sin(4π/7) = 4cs(2c-1)

    Since sin(3π/7) = sin(4π/7), we have 4c - 1 = 4c(2c - 1) or

    (1): 8c = 4c + 4c - 1

    This means any polynomial in c can be reduced to a quadratic or less.

    However, it will take a bit of work to simplify the still messy expression of

    P = 2cs(4c-1)

    = s(8c - 2c) and now apply (1) to get

    = s(4c + 2c - 1)

    Now we take a slight detour:

    (2): s(4c + 2c - 1)

    = (1-c)(4c + 2c - 1)

    = -4c⁴ -2c + 5c + 2c -1 and now apply (1)

    = -4c + 3c + (5/2)c - 1 and apply (1) again to get

    = c + c -

    We now return to our regularly scheduled programming:

    P = (s(4c + 2c - 1)) apply (2)

    = (s(c + c/2 - 1/2))

    = s(c + c - - )(c + c - ) apply (2) again

    = ((c + c - ) - (1-c)) * (c + c - )

    = (4c + c - 3) * (2c + c -1) / 16

    = (8c⁴ + 6c - 9c - 4c + 3) / 16 apply (1)

    = (10c - 5c - 5c + 3) / 16 apply (1) again

    = (-5/4 + 3) / 16

    = (7/4) / 16

    Hence P = ⅛ * √7

    Remark: In the other question referenced, we'd already seen that s(4c +2c -2)= *√7, so we could have avoided this last section of calculation by observing that:

    P = s(c + c/2 - 1/2) = *s(4c + 2c - 2) = ⅛ * √7

    We've done!

    2009-05-25 00:18:28 補充:

    You are looking for P = sin(π/7)sin(2π/7)sin(3π/7)

    Let s = sin(π/7), c = cos(π/7) so s² + c² = 1.

    From cos(α + β) and sin(α + β) formulas come

    cos(2π/7) = c² - s² = 2c² - 1

    sin(2π/7) = 2cs

    sin(3π/7) = s(2c² - 1) + c(2cs) = s(4c² - 1), and

    sin(4π/7) = 4cs(2c²-1)

    2009-05-25 00:21:02 補充:

    Since sin(3π/7) = sin(4π/7), we have 4c² - 1 = 4c(2c² - 1) or

    (1): 8c³ = 4c² + 4c - 1.This means any polynomial in c can be reduced to a quadratic or less.

    2009-05-25 00:21:12 補充:

    However, it will take a bit of work to simplify the still messy expression of

    P = 2cs³(4c²-1)= s³(8c³ - 2c) and now apply (1) to get

    = s³(4c² + 2c - 1)

    2009-05-25 00:21:38 補充:

    Now we take a slight detour:

    (2): s²(4c² + 2c - 1)= (1-c²)(4c² + 2c - 1)

    = -4c⁴ -2c³ + 5c² + 2c -1 and now apply (1)

    = -4c³ + 3c² + (5/2)c - 1 and apply (1) again to get

    = c² + ½c - ½

    2009-05-25 00:22:22 補充:

    We now return to our regularly scheduled programming:

    P² = (s³(4c² + 2c - 1))² apply (2)

    = (s(c² + c/2 - 1/2))²= s²(c² + ½c - ¼ - ¼)(c² + ½c - ½) apply (2) again

    = (¼(c² + ½c - ½) - ¼(1-c²)) * (c² + ½c - ½)

    2009-05-25 00:22:54 補充:

    hence P²= (4c² + c - 3) * (2c² + c -1) / 16

    = (8c⁴ + 6c³ - 9c² - 4c + 3) / 16 apply (1)

    = (10c³ - 5c² - 5c + 3) / 16 apply (1) again

    = (-5/4 + 3) / 16= (7/4) / 16

    2009-05-25 00:23:16 補充:

    Hence P = ⅛ * √7

    Remark: In the other question referenced, we'd already seen that s(4c² +2c -2)= ½*√7, so we could have avoided this last section of calculation by observing that:P = s(c² + c/2 - 1/2) = ¼*s(4c² + 2c - 2) = ⅛ * √7

  • YO
    Lv 5
    1 decade ago

    sin3x = sin4x 有四根 x=π/7、3π/7、5π/7、7π/7

    sin3x = -sin4x 有四根 x=0、2π/7、4π/7、6π/7

    因此 (sin4x)^2-(sin3x)^2=0對s展開後,約掉s^2項,由根與係數

    即可得A^2

    => 16(1-s^2)(1-2s^2)^2-(3-4s^2)^2=0

    領導係數-16*2^2=-64,常數項16-3^2=7

    => A^2 = 7/64

  • linch
    Lv 7
    1 decade ago

    不知道到底 Yahoo 支援哪些字體?

  • 1 decade ago

    Great!

    請問c⁴⅛中4次方與分數怎麼打出來的?

    2009-05-25 01:14:20 補充:

    找到MS Mincho字體了!

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