Trigonometry
聽日要考啦,有幾條exercise唔多識做!1)
Show that sinx+sin2x+sin3x+...+sinnx={sin(1/2 nx)sin}/sin(1/2 x) for allpositive integers n
2)
Express cos^n x in terms of multiple angles, for positive integer n.
3)
Express sin(nx) in terms of sinx and cosx, for positive integer n.
(第一題最好唔好用MI啦,我想睇下有冇其他方法,希望有人答到我!) 本帖最後由 -終場ソ使者- 於 4-11-2010 18:23 編輯
岩岩考完試@@
1)
Let http://upload.wikimedia.org/math/0/1/9/0190d0aea4c30829ff0fdbb9f74e89a6.png and http://upload.wikimedia.org/math/c/9/6/c962cd4425252358befbaef3c2870f82.png,
http://upload.wikimedia.org/math/2/4/0/24088e0795600c523e060b29a98acd16.png
http://upload.wikimedia.org/math/6/b/7/6b7da18d656c6adcf1803a54fbe3b38d.png
http://upload.wikimedia.org/math/8/d/6/8d69578a71d4f68ea5d148e76f219eb0.png (geometric series)
http://upload.wikimedia.org/math/7/0/8/70884dbb37f780004ae47f0921de3d47.png
http://upload.wikimedia.org/math/4/8/b/48b445bfbbb4df871b12844011ae6946.png
http://upload.wikimedia.org/math/1/6/7/1676b99f1eebed58cff4a3dc52085f4a.png
http://upload.wikimedia.org/math/0/f/c/0fc5dade9e79608eb8e70c209c9bc05d.png
Hence, http://upload.wikimedia.org/math/5/9/7/597ae4bc16f5274c130ae6fc9b3c8b43.png
(PS: an extra series for cosine can be obviously found as http://upload.wikimedia.org/math/5/5/8/558e089d4df390e7b230e3f9a5bb4a4d.png)
2)
Since http://upload.wikimedia.org/math/9/b/3/9b3d4229d0f74b02dd5dc019c642d99f.png
http://upload.wikimedia.org/math/0/8/d/08d70923c8adc494c3b2d8a717d9a9f4.png
http://upload.wikimedia.org/math/4/3/4/434b161dcc24f08ae044974b466f9179.png
http://upload.wikimedia.org/math/f/4/f/f4fb213d923a41be27821f49af23ac1e.png
Hence, http://upload.wikimedia.org/math/7/6/1/761c97a638c66b7b13cb62cc78f95179.png
呢條唔多肯定,
3)(諗緊,一陣先做) 我唔多明第3條想問咩 有無中文翻譯? 求sin nx,答案以sinx,cosx表示
eg: sin2x=2sinxcosx 本帖最後由 p445hkk20001 於 4-11-2010 04:49 編輯
sin3X = 3sinX-4sin^3 (X)
好難寫關係@@''
順帶一提 你個簽名 即係334個網條LINK係死 third question use de moivre's theorem? 本帖最後由 -終場ソ使者- 於 5-11-2010 18:06 編輯
咁就用De Moivre's thm@@
http://upload.wikimedia.org/math/6/e/0/6e05509c9c0cb919f4a4054e43db4c49.png
http://upload.wikimedia.org/math/a/b/2/ab243ed2f7278ac16b698b81a1a8b738.png 本帖最後由 p445hkk20001 於 6-11-2010 06:20 編輯
For the sake of convenience,we set C =cosθ S=sinθ
by themoivre's theorem,
(C+iS)^n=cosnθ+isinnθ
The binomial expansion of the left hand side is
(C+iS)^n
=C^n+nC1 (C^n-1)(iS)+(nC2)(C^n-2)(iS)^2+(nC3)(C^n-3)(iS)^3....+....
+(nCn-1)C x (iS)^n-1 +(iS)^n
={C^n - (nC2)C^(n-2)S^2+(nC4)C^(n-4)S^4 - .......}
+i{(nC1)C^(n-1)S-(nC3)C^(n-3)S^3+(nC5)C^(n-5)S^5 -........}
equating real and imaginary parts of both sides ,we get
sinnθ=(nC1)C^(n-1)S-(nC3)C^(n-3)S^3+(nC5)C^(n-5)S^5 -........(可能錯)
話時話m2 會考咁多i既野咩?
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