化简sinX^2+cosX^2的具体步骤~~~ sinxcosx+cosx^2-2 化简

sinx^2cosx^2\u7684\u5316\u7b80\u63a8\u5bfc\u8fc7\u7a0b

sin^2xcos^2x=1/4sin^2(2x)=1/8*\uff081-cos4x\uff09
\uff08sinx\uff09^2\uff08cosx\uff09^2=\uff081/4\uff09\uff082sinxcosx)^2
=(1/4)(sin2x)^2=(1/4)*(1/2)(1-cos4x)
=(1/8)(1-cos4x)
sin²xcos²x
=(1/4)[2sinxcosx]²
=(1/4)sin²2x
=(1/8)[2sin²x]
=(1/8)[1\uff0dcos4x]
=(1/8)\uff0d(1/8)cos4x
\u4e09\u89d2\u51fd\u6570
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sinxcosx+cosx^2-2
=(1/2)*2sinxcosx+(1/2)*[2*(cosx)^2-1]-(3/2)
=(1/2)sin2x+(1/2)cos2x-(3/2)
=(\u221a2/2)sin(2x+\u03c0/4)-(3/2)

f(x)=(1 sinx cosx)[sin(x/2)-cos(x/2)]/√（2 2cosx） =[sin(x/2) cos(x/2) sinx cosx][sin(x/2)-cos(x/2)]/[√2（1 cosx）] =[sin(x/2) cos(x/2) 2sin(x/2)*cos(x/2) cos(x/2)-sin(x/2)][sin(x/2)-cos(x/2)]/[√4*（1 cosx）/2] =｛[sin(x/2) cos(x/2)] [cos(x/2) sin(x/2)][cos(x/2)-sin(x/2)]｝[sin(x/2)-cos(x/2)]/[√4*cos(x/2)] =｛[sin(x/2) cos(x/2)][sin(x/2) cos(x/2） cos(x/2)-sin(x/2)]｝[sin(x/2)-cos(x/2)]/[√4*cos(x/2)] =｛[sin(x/2) cos(x/2)]*2cos(x/2）｝[sin(x/2)-cos(x/2)]/[2*cos(x/2)] =｛[sin(x/2) cos(x/2)][sin(x/2)-cos(x/2)]*2cos(x/2）｝/[2*cos(x/2)] =[sin(x/2) cos(x/2)][sin(x/2)-cos(x/2)] =sin(x/2)-cos(x/2) =-cosx

sinx^2cosx^2鐨鍖栫畝鎺ㄥ杩囩▼
绛旓細锛坰inx锛塣2锛坈osx锛塣2=锛1/4锛锛2sinxcosx)^2 =(1/4)(sin2x)^2=(1/4)*(1/2)(1-cos4x)=(1/8)(1-cos4x)sin²xcos²x =(1/4)[2sinxcosx]²=(1/4)sin²2x =(1/8)[2sin²x]=(1/8)[1锛峜os4x]=(1/8)锛(1/8)cos4x 涓夎鍑芥暟 涓夎鍑芥暟...

sinx^2cosx^2鐨勪笉瀹氱Н鍒嗘槸浠涔?
绛旓細sinx^2cosx^2 =[(sin2x)/2]^2 =[(sin2x)^2]/4 =(1-cos4x)/8.锛坰inx^2cosx^2锛=(1/8)[x-(sin4x)/4]+C =x/8-(sin4x)/32+C 鎵浠inx^2cosx^2鐨勪笉瀹氱Н鍒嗘槸x/8-(sin4x)/32+C銆

sin^2cos^2鍖栫畝
绛旓細sin^2cos^2=(cos2a+1)/2*(1-cos2a)/2=1-cos2a^2/4=sin2a^2/4

(sinx)^2*(cosx)^2鐨勪笉瀹氱Н鍒嗘庝箞姹傚憿
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sinx^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...

(sinx)^2(cosx)^2鐢ㄤ簩鍊嶈鍏紡鎬庝箞鍖栫畝鍑烘潵(sin2x)^2/4鐨?
绛旓細2sinxcosx=sin2x sinx^2cosx^2=sin2x/2 * sin2x/2=锛坰in2x)^2/4

姹備笉瀹氱Н鍒(sinx^2cosx^2)
绛旓細sinx^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

鈭sinx^2.cosx^2 dx,
绛旓細sinx*sinx*cosx*cosx=sin2x*sin2x/4,鈭sinx^2.cosx^2 dx=鈭玸in2x*sin2x/4 dx =-1/8鈭玸in2x dcos2x =鈥︹︿笅闈㈠氨绠鍗曚簡

y=sinx^2涔樹互cosx^2,姹傚
绛旓細鏂规硶涓锛歽=(sinx)^2路(cosx)^2,y'=[(sinx)^2]'路(cosx)^2+(sinx)^2路[(cosx)^2]'=2sinx路(sinx)'路(cosx)^2+(sinx)^2路2cox路(cosx)'=2sinx(cosx)^3-2(sinx)^3cosx 澶囨敞锛 鍖栫畝鐪嬬湅 =2sinxcosx[(cox)^2-(sinx)^2]=sin2x路cos2x =1/2路2sin2xcos2x =1/2 路sin...

宸茬煡f鈥榹(sinx)^2}=(cosx)^2姹俧(x)
绛旓細瑙ｇ瓟濡備笅锛歠鈥榹锛sinx锛^2}=锛cosx锛塣2锛堟渶鍓嶉潰鐨勯偅鏄鏁板悧锛燂級f鈥榹锛坰inx锛塣2}=1-锛坰inx锛塣2 鎶婏紙sinx锛塣2褰撴垚涓涓暣浣撳氨鍙緱 f鈥橈紙x锛=1-x 濡傛灉娌℃湁瀵兼暟锛岄偅涔堝氨鏄痜锛坸锛=1-x 濡傛灉鏈夊鏁扮殑璇濓紝閭ｄ箞f锛坸锛=x-(1/2)x^2+a(a涓哄父鏁)~璇烽鍏堝叧娉ㄣ愭垜鐨勯噰绾崇巼銆憕濡傛灉浣犺鍙...

sinx^2+cosx^2绛変簬浠涔?
绛旓細锛坰inx锛锛2锛坈osx锛夛季2 锛濓紙1/4锛夛紙2sinxcosx)^2 =(1/4)(sin2x)^2 =(1/4)*(1/2)(1-cos4x)=(1/8)(1-cos4x)sinx鐨勬渶鍊煎拰闆剁偣 鈶犳渶澶у硷細褰搙=2k蟺+(蟺/2) 锛宬鈭圸鏃讹紝y(max)=1 鈶℃渶灏忓硷細褰搙=2k蟺+(3蟺/2)锛宬鈭圸鏃讹紝y(min)=-1 闆跺肩偣锛 (k蟺,0) 锛宬鈭圸...

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