SIN2X的平方=1－COS4X的平方/2是怎么演变而来的
\uff08sin3xsinx\uff09sin2x+\uff08cos3xcosx\uff09cos2x =1/ 2 [\uff08cos2x-cos4x\uff09sin2x+\uff08cos2x+cos4x\uff09cos2x] \u662f\u600e\u4e48\u53d8\u7684\u79ef\u5316\u548c\u5dee\u516c\u5f0f
sinasinb=1/2[cos(a-b)-cos(a+b)]
cosacosb=1/2[cos(a-b)+cos(a+b)]
\u89e3
f(x)=(2cos²-1)sin2x+1/2cos4x
=cos2xsin2x+1/2cos4x
=1/2sin4x+1/2cos4x
=\u221a2/2sin(4x+\u03c0/4)
\u2234
T=2\u03c0/4=\u03c0/2\u4e3a\u5176\u6700\u5c0f\u5468\u671f
\u6700\u5927\u503c\u4e3a\uff1a\u221a2/2
f(a)=\u221a2/2sin(4a+\u03c0/4)=\u221a2/2
\u22344a+\u03c0/4=\u03c0/2+2k\u03c0
\u2234a=\u03c0/16+k\u03c0/2
\u2235a\u2208(\u03c0/2,\u03c0)
\u2234k=1\u65f6\uff0ca=9\u03c0/16\u2208(\u03c0/2,\u03c0)
\u2234a=9\u03c0/16
1=sin^2 2x+cos^2 2x
右边=2sin^2 2x/2=sin^2 2x=左边
SIN2X的平方+COS2X的平方=1
COS4X=COS2X的平方-SIN2X的平方
联立则是 2（SIN2X的平方）=1-COS4X
绛旓細鍏堢粰浣犺涓や釜鍏紡锛氫竴锛COS2x=1-2(sinx)鐨勫钩鏂銆sin2x=2sinxcosx銆2-2COS8=4(sin4)鐨勫钩鏂广傚紑鏂瑰悗涓猴細2sin4銆1-sin8=(sin4-cos4)鐨勫钩鏂广傚紑鏂瑰悗涓猴細sin4-cos4銆傜粨鏋滀负锛4sin4-2cos4銆
绛旓細鏈鍩烘湰鐨勫嶈鍑芥暟鍏紡鍟婏紒cost=1-2sin²(t/2)鎵浠ワ紝1-cost=2sin²(t/2)鎵浠ワ紝(1-cost)^4=[2sin²(t/2)]^4=16sin^8 (t/2)
绛旓細2脳(sinx)^2銆傜敱鍏紡锛氬彲寰楋細cos2x=2*(cosx)^2-1=1-2*(sinx)^2 鏁咃細1-cos2x=1-[1-2*(sinx)^2]=2脳(sinx)^2銆傚嶈鍏紡锛屾槸涓夎鍑芥暟涓潪甯稿疄鐢ㄧ殑涓绫诲叕寮忋傚氨鏄妸浜屽嶈鐨勪笁瑙掑嚱鏁扮敤鏈鐨勪笁瑙掑嚱鏁拌〃绀哄嚭鏉ャ傚湪璁＄畻涓彲浠ョ敤鏉ュ寲绠璁＄畻寮忋佸噺灏戞眰涓夎鍑芥暟鐨勬鏁帮紝鍦ㄥ伐绋嬩腑涔熸湁骞挎硾鐨...
绛旓細鍏蜂綋鍥炵瓟濡備笅锛cos鐨勫洓娆℃柟+sin鐨勫洓娆℃柟 =锛坰in伪鐨勫钩鏂+cos伪鐨勫钩鏂癸級鐨勫钩鏂-4sin伪鐨勫钩鏂*cos伪鐨勫钩鏂 =1-4sin伪鐨勫钩鏂*cos伪鐨勫钩鏂 =1-锛2sin伪*cos伪锛夌殑骞虫柟 =1-2sin2伪鐨勫钩鏂 =cos4伪 鍜岃鍏紡锛歴in ( 伪 卤 尾 ) = sin伪 路 cos尾 卤 cos伪 路 sin尾 sin ( 伪 + 尾...
绛旓細鎵浠ュ緱鍒(sinx)^2 *(cosx)^2銆=1/4 *(2sinxcosx)^2銆=1/4 *(sin2x)^2銆傚崟浣嶆崲绠楋細1 銕★紙1骞虫柟绫筹級= 100 dm²锛100骞虫柟鍒嗙背锛=10000 cm²锛10000骞虫柟鍘樼背锛=1000000 mm²锛1000000骞虫柟姣背锛= 0.0001鍏》=0.000001km² 锛0.000001骞虫柟鍏噷锛= 0.01鍏憨=0....
绛旓細sin²伪+cos²伪=1 sin鐨勫钩鏂銆乧os鐨勫钩鏂广 tan鐨勫钩鏂 鐨勫叕寮忔槸锛1銆乻in²伪+cos²伪=1 2銆1+tan²伪=sec²伪 3銆1+cot²伪=csc²伪 4銆乻in²伪=(1-cos2a)/2 5銆乧os²a=(1+cos2a)/2 6銆乼an²a=(2tana-1)/(tan2a...
绛旓細鍏蜂綋鍥炵瓟濡備笅锛氾紙sin2x锛²'=2sin(2x) *[sin(2x)]'=2sin(2x)cos2x*(2x)'=4sin(2x)cos(2x)=2sin(4x)瀵兼暟鐨勬剰涔夛細瀵逛簬鍙鐨勫嚱鏁癴(x)锛寈↦f'(x)涔熸槸涓涓嚱鏁帮紝绉颁綔f(x)鐨勫鍑芥暟锛堢畝绉板鏁帮級銆傚鎵惧凡鐭ョ殑鍑芥暟鍦ㄦ煇鐐圭殑瀵兼暟鎴栧叾瀵煎嚱鏁扮殑杩囩▼绉颁负姹傚銆傚疄璐ㄤ笂锛屾眰瀵煎氨鏄竴...
绛旓細2sinAcosB=sin(A+B)+sin(A-B)2cosAsinB=sin(A+B)-sin(A-B) )2cosAcosB=cos(A+B)-sin(A-B)-2sinAsinB=cos(A+B)-cos(A-B)sinA+sinB=2sin((A+B)/2)cos((A-B)/2 cosA+cosB=2cos((A+B)/2)sin((A-B)/2)tanA+tanB=sin(A+B)/cosAcosB 鏌愪簺鏁板垪鍓峮椤瑰拰 1+2+3+4...
绛旓細(sinx)^2*(cosx)^2鐨勪笉瀹氱Н鍒嗘槸x/8-(sin4x)/32+C銆傝В锛歴inx^2cosx^2 =[(sin2x)/2]^2 =[(sin2x)^2]/4 =(1-cos4x)/8 涓嶅畾绉垎锛坰inx^2cosx^2锛=(1/8)[x-(sin4x)/4]+C=x/8-(sin4x)/32+C 鎵浠(sinx)^2*(cosx)^2鐨勪笉瀹氱Н鍒嗘槸x/8-(sin4x)/32+C銆備笉瀹氱Н鍒嗙殑...
绛旓細cosx =1-2sin(x/2)^2 1-cosx=2sin(x/2)^2 鐢变簬x瓒嬩簬0,鍒檟/2瓒嬩簬0,sin(x/2)鍜(x/2)绛変环 1-cosx=2*锛坸/2)^2 =x^2/2
