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請教一題三角函數

求\( \displaystyle \root 3 \of {cos\frac{2 \pi}{9}cos \frac{4 \pi}{9}}+\root 3 \of {cos\frac{4 \pi}{9}cos \frac{8 \pi}{9}}+\root 3 \of {cos\frac{8 \pi}{9}cos \frac{2 \pi}{9}} \)的值

求以\( \displaystyle cos \frac{2 \pi}{9}, cos \frac{4 \pi}{9},cos \frac{8 \pi}{9} \)為根的三次方程式
\( \displaystyle \theta=\frac{2 \pi}{9},\frac{4 \pi}{9},\frac{8 \pi}{9}, \)滿足\( cos 3\theta=-\frac{1}{2} \)
\( cos3 \theta=4cos^3 \theta-3cos \theta \)
所求方程式為\( \displaystyle 4x^3-3x=-\frac{1}{2} \) , \( 8x^3-6x+1=0 \)
利用根與係數的關係可以得到三根之和為0,兩兩相乘的和為\( \displaystyle -\frac{3}{4} \),三根之積為\( \displaystyle -\frac{1}{8} \)


若要求三根平方的方程式
\( x(8x^2-6)=-1 \) , \( x^2(8x^2-6)^2=1 \) , \( y(8y-6)^2=1 \) , \( 64y^3-96y^2+36y-1=0 \)
若要求三根立方的方程式
\( 8x^3+1=6x \) , \( (8x^3+1)^3=(6x)^3 \) , \( (8y+1)^3=216y \) , \( 512y^3+192y^2-192y+1=0 \)
但這題要的是立方根的方程式
\( 8x^3+1=6x \) , \( \root 3 \of {8x^3+1}=\root 3 \of {6x}  \)
將\( \root 3 \of {x} \)換成\( y \),但左式就換不出來了

或許有其他的方法可以解出答案

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我在淡江圖書館找到這題的解答
Jiří Herman, Radan Kučera, Jaromír Šimša
Equations and Inequalities:Elementary Problems and Theorems in Algebra and Number Theory
第81頁

書上的題目是\( \displaystyle \root 3 \of{cos \frac{2 \pi}{7}}+\root 3 \of{cos \frac{4 \pi}{7}}+\root 3 \of{cos \frac{6 \pi}{7}}=\root 3 \of{\frac{1}{2}(5-3 \root 3 \of 7)} \)
相同的方法就可以解出答案
\( \displaystyle \root 3 \of {cos \frac{2 \pi}{9}}+\root 3 \of{cos \frac{4 \pi}{9}}+\root 3 \of{cos \frac{8 \pi}{9}}=\root 3 \of{\frac{3}{2}\root 3 \of 9-3} \)
\( \displaystyle \root 3 \of{cos \frac{2 \pi}{9}cos \frac{4 \pi}{9}}+\root 3 \of{cos \frac{4 \pi}{9}cos \frac{8 \pi}{9}}+\root 3 \of{cos \frac{8 \pi}{9}cos \frac{2 \pi}{9}}=\root 3 \of{\frac{3}{4}(1-\root 3 \of 9)} \)

用maxima驗證答案
https://math.pro/db/viewthread.php?tid=709&page=3#pid2756

109.7.12補充
\(\root 3\of{cos40^{\circ}}+\root 3\of{cos80^{\circ}}-\root 3\of{cos20^{\circ}}=\root 3\of{\displaystyle \frac{3}{2}(\root 3 \of{9}-2)}\)
(101高中數學能力競賽,h ttp://teach.nehs.tc.edu.tw/downFile.php?file=download/unitfile/7/file0_20170221115958.pdf&filename=101%E5%AD%B8%E5%B9%B4%E5%BA%A6%E9%AB%98%E4%B8%AD%E6%95%B8%E5%AD%B8%E8%83%BD%E5%8A%9B%E7%AB%B6%E8%B3%BD%E5%85%A8%E5%9C%8B%E6%B1%BA%E8%B3%BD.pdf連結已失效)

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補上相關的題目
Prove the identity
\( \displaystyle \root 3 \of {cos \frac{2 \pi}{7}}+\root 3 \of{cos \frac{4 \pi}{7}}+\root 3 \of{cos \frac{8 \pi}{7}}=\root 3 \of{\frac{1}{2}(5-3\root 3 \of 7)} \)

https://artofproblemsolving.com/community/c6h370998
https://artofproblemsolving.com/community/c6h363946p1998666
https://artofproblemsolving.com/community/c6h1079891p4739612



\( \displaystyle \omega=cos \frac{2 \pi}{7}+i sin \frac{2 \pi}{7} \)
\( \displaystyle \alpha=\omega+\omega^6=2 cos \frac{2 \pi}{7} \),
\( \displaystyle \beta=\omega^2+\omega^5=2 cos \frac{4 \pi}{7} \),
\( \displaystyle \gamma=\omega^3+\omega^4=2 cos \frac{6 \pi}{7} \),
求以\( \alpha,\beta,\gamma \)為三根的三次方程式為?
(88高中數學能力競賽 宜花東區試題)
h ttp://www.math.nuk.edu.tw/senpengeu/HighSchool/2000_Taiwan_High_Ilan_02.pdf連結已失效


試證:\( \displaystyle 2 cos \frac{2 \pi}{7} \)為\( x^3+x^2-2x-1 \)之一根
(88高中數學能力競賽 高屏區試題)
h ttp://www.math.nuk.edu.tw/senpengeu/HighSchool/2000_Taiwan_High_Pingtung_02.pdf連結已失效


若\( \displaystyle \frac{n}{100}<2 cos \frac{2 \pi}{7}<\frac{n+1}{100} \),\( n \in N \),則n=。
(99建國中學,https://math.pro/db/thread-968-1-1.html)

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書裡還有很多值得準備的題目,google books有的我就將網址列出來
沒有的我就將內容抄出來,想知道更多題目可以到圖書館去找這本書。

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Prove that the number \( \displaystyle c=\root 3 \of {\frac{1}{9}}+\root 3 \of {-\frac{2}{9}}+\root 3 \of {\frac{4}{9}} \) is a zero of \( F(x)=x^3+\root 3 \of 6 x^2-1 \).
第44頁
http://books.google.com.tw/books?id=l0CDp-YVyjoC&lpg=PP1&ots=01zdioOshi&dq=Equations%20and%20Inequalities:Elementary%20Problems%20and%20Theorems%20in%20Algebra%20and%20Number%20Theory&pg=PA44#v=onepage&q&f=true



求滿足下列等式的數a:\( \displaystyle \root 3 \of{\root 3 \of 2 -1}=\frac{1}{\root 3 \of a}(1-\root 3 \of 2+\root 3 \of 4) \)
(1995日本奧林匹克預選賽)

若\( \displaystyle \root 3 \of{\root 3 \of 2 -1}=\root 3 \of a+\root 3 \of b+\root 3 \of c \),則\( a+b+c= \)?
(92高中數學能力競賽,高中數學101 P25)

110.9.23補充
\( \displaystyle \root 3 \of{\root 3 \of 2 -1}=\root 3 \of a+\root 3 \of b+\root 3 \of c \),其中\(a\)、\(b\)、\(c \in Q\)。求\(a+b+c=\)   
(100中科實中,https://math.pro/db/viewthread.php?tid=1107&page=3#pid3377)

當\(\root 3\of{\root 3\of{16}-2}=\root 3\of a+\root 3\of b+\root 3\of c\),其中\(a,b,c \in Q\)。求\(a+b+c\)。
(109第1學期中山大學雙週一題)

110.2.25補充
若\(a\)是一個有理數且滿足\(\displaystyle \frac{1}{\root 3 \of 4+\root 3 \of 2+a}=\alpha \root 3 \of 4+\beta \root 3 \of 2+\gamma\),其中\(\alpha,\beta,\gamma\)為有理數。試求\(\alpha,\beta,\gamma\)(用\(a\)表示)
(109北科附工,https://math.pro/db/viewthread.php?tid=3326&page=2#pid21223)

111.1.10補充
有理化\( \displaystyle \frac{1}{\root 3 \of 2+\root 3 \of 3+\root 3 \of 5} \)
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For the expression
\( \displaystyle Q(n)=\frac{1}{2}\cdot \frac{3}{4}\cdot \frac{5}{6}\cdot \cdot \cdot \frac{2n-1}{2n} \)        ( \( n \ge 2 \) )
we will prove the bounds \( \displaystyle \frac{1}{2 \sqrt{n}}<Q(n)<\frac{1}{\sqrt{2n+1}} \).
第103頁
http://books.google.com.tw/books?id=l0CDp-YVyjoC&lpg=PP1&ots=01zdioOshi&dq=Equations%20and%20Inequalities:Elementary%20Problems%20and%20Theorems%20in%20Algebra%20and%20Number%20Theory&pg=PA103#v=onepage&q&f=true


\( \displaystyle a_n=\frac{1 \cdot 3 \cdot 5 \cdot \cdot \cdot (2n-1)}{2 \cdot 4 \cdot 6 \cdot \cdot \cdot (2n)} \),試求\( \displaystyle \lim_{n \to \infty}a_n \)。
(97文華高中,http://forum.nta.org.tw/examservice/showthread.php?t=47781)

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Show that for arbitary \( x,y,z \in R \) we have
\( \displaystyle (\frac{x}{2}+\frac{y}{3}+\frac{z}{6})^2 \le \frac{x^2}{2}+\frac{y^2}{3}+\frac{z^2}{6} \)        (43)
Furthermore,determine when equality occures in (43).
第129頁
http://books.google.com.tw/books?id=l0CDp-YVyjoC&lpg=PP1&ots=01zdioOshi&dq=Equations%20and%20Inequalities:Elementary%20Problems%20and%20Theorems%20in%20Algebra%20and%20Number%20Theory&pg=PA129#v=onepage&q&f=true


求\( \displaystyle (\frac{x}{2}+\frac{y}{3}+\frac{z}{6})^2=\frac{x^2}{2}+\frac{y^2}{3}+\frac{z^2}{6} \)的所有整數解。
(90高中數學能力競賽 中彰投區試題)
h ttp://www.math.nuk.edu.tw/senpengeu/HighSchool/2002_Taiwan_High_Taichung_02.pdf 連結已失效


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Show that the inequality \( \sqrt{x+1}+\sqrt{2x-3}+\sqrt{50-3x}\le 12 \) holds for all values \( x \in R \) for which the left-hand side is defined.

SOLUTION. By (48) with \( n=3 \), \( u_1=\sqrt{x+1} \),\( u_2=\sqrt{2x-3} \), and \( u_3=\sqrt{50-3x} \)
we have \( (\sqrt{x+1}+\sqrt{2x-3}+\sqrt{50-3x})^2 \le 3(x+1+2x-3+50-3x)=144 \),
from which, upon taking the square root, we obtain the result.


求函數\( y=\sqrt{x+27}+\sqrt{13-x}+\sqrt{x} \)的最大和最小值。
(2009大陸高中數學競賽)


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Given an aribitrary prime p, solve the Diophantine equation \( x^4+4^x=p \).(第245頁)
SOLUTION. For any integer \( x<0 \), \( 4^x \) is not an integer, and neither is \( x^4+4^x \).
Thus for any prime p, the given equation has no negative solution.
For \( x=0 \) we have \( 0^4+4^0=1 \), which is not a prime, and for \( x=1 \) we have \( 1^4+4^1=5 \), a prime.
We will now show that for any integer \( x\ge 2 \), the number \( x^4+4^x \) is composite. If \( x=2k \) is even, where \( k \in N \), then
\( x^4+4^x=2^4 k^4+4^{2k}=16(k^4+4^{2(k-1)}) \),
which is a composite number. If \( x=2k+1 \)( \( k \in N \) ) is odd, then
\( x^4+4^x=x^4+4 \cdot 4^{2k}=[x^4+4x^2(2^k)^2+4(2^k)^4]-4x^2(2^k)^2 \)
    \( =[x^2+2(2^k)^2]^2-(2x \cdot 2^k)^2 \)
    \( =[x^2+2x \cdot 2^k+2(2^k)^2][x^2-2x \cdot 2^k+2(2^k)^2] \)
    \( =[(x+2^k)^2+2^{2k}][(x-2^k)^2+2^{2k}] \),
which is again compostie, since \( (x±2^k)^2+2^{2k}\ge 2^{2k} \ge 2^2 >1 \).
To summarize, for \( p=5 \) the given equation has the unique solution \( x=1 \),while there are no solutions for any prime \( p \ne 5 \)


Prove that if \( n>1 \) then \( n^4+4^n \) is compostie.
http://www.artofproblemsolving.com/Forum/viewtopic.php?t=328143


設\( n,a \)為自然數,試證:有無限多個a使得\( n^4+a \)必不為質數
(藍藍天上一朵雲 歷屆教甄考題整理2005版第75題)


Prove that there are infinitely many natural numbers a with the following property: the number \( z=n^4+a \) is not prime for any natural number n.
(1969IMO,http://www.artofproblemsolving.c ... p/1969_IMO_Problems)


Compute \( \displaystyle \frac{(10^4+324)(22^4+324)(34^4+324)(46^4+324)(58^4+324)}{(4^4+324)(16^4+324)(28^4+324)(40^4+324)(52^4+324)} \).
(1987AIME)
101.1.10補充
100卓蘭實驗高中,https://math.pro/db/thread-1165-1-1.html
高中數學101 P26

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