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#1
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ªèǾÔÊÙ¨¹ì˹èͤÃѺ
1. 1 + cosA + cos2A + cos 3A = 4 cosA.cos3/2A.cosA/2
2. cos^3 A.cos3A + sin^3 A.sin 3A = cos^3 2A ¢Íº¤Ø³¤ÃѺ¼Á |
#2
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¢éÍáá¡è͹¹Ð¤ÃѺ
Êٵà cos 2A = 2cos^2 A - 1 cos 3A = 4cos^3 A - 3cos A á·¹¤èÒ 1 + cos A + 2cos^2 A - 1 + 4cos^3 A - 3cos A = 4cos^3 A + 2cos^2 A - 2cos A = 2cos A ( 2cos^2 A + cos A - 1) á·¹¤èÒ cos^2 A = (1 + cos 2A )/ 2 ´Ñ§¹Ñé¹ = 2cos A( 2(1 + cos 2A)/2 + cos A - 1 ) = 2cos A( 1 + cos2A + cos A - 1 ) = 2cos A (cos A + cos2A) Êٵà cos A + cos B = 2cos(A+B)/2 cos(A-B)/2 ´Ñ§¹Ñé¹ cosA + cos2A = 2cos(3A/2)cos(A/2) á·¹¤èÒ ¨Ðä´é 2cos A [ 2cos(3A/2)cos(A/2) ] = 4 cos (A/2) cos A cos (3A//2) «.µ.¾. |
#3
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¢éÍ 1.
1+cosA+cos2A+cos3A =(cos0+cos3A)+(cosA+cos2A) =2cos(3/2)Acos(3/2)A+2cos(3/2)AcosA/2 =2cos(3/2)A[cos(3/2)A+cosA/2] =2cos(3/2)A[2cosAcosA/2] =4cosAcos(3/2)AcosA/2 ¢éÍ 2. à¹×èͧ¨Ò¡ cos3A=4(cosA)^3-3cosA sin3A=3sinA-4(sinA)^3 ¨Ðä´é (cosA)^3=(3cosA+cos3A)/4 (sinA)^3=(3sinA-sin3A)/4 ´Ñ§¹Ñé¹ (cosA)^3(cos3A)+(sinA)^3(sin3A) =[3cosAcos3A+(cos3A)^2+3sinAsin3A-(sin3A)^2]/4 =[3(cosAcos3A+sinAsin3A)+(cos3A)^2-(sin3A)^2]/4 =[3cos2A+cos6A]/4 =(cos2A)^3 |
#4
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1
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#5
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#6
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