y=sin的4次方x+cos的2次方x 化简最后是多少。化成统一个三角函数 已知函数f(x)=cos的4次方x+2sinxcosx--s...
\u6c42\u8bc1\u660e\uff1asin\u76844\u6b21\u65b9x\uff0bcos\u76844\u6b21\u65b9x=1-2sin\u76842\u6b21\u65b9x\u4e58cos\u76842\u6b21\u65b9x(sinx^2)^2+(cosx^2)^2-2sinx^2cosx^2+2sinx^2cosx^2
(sinx^2+cosx^2)^2-2sinx^2cosx^2
\u56e0\u4e3acosx^2=\uff081+cos2x\uff09/2\uff0c sinx^2=\uff081-cos2x\uff09/2
sin^2+cos^2=1
\u6240\u4ee51-2sinx^2cosx^2
f(x)=cos\u76844\u6b21\u65b9x-sin\u76844\u6b21\u65b9x+2sinxcosx
=(cos\u76842\u6b21\u65b9x-sin\u76842\u6b21\u65b9x)(cos\u76842\u6b21\u65b9x+sin\u76842\u6b21\u65b9x)+sin2x
=cos\u76842\u6b21\u65b9x-sin\u76842\u6b21\u65b9x+sin2x
\u2235cos2x=cos\u76842\u6b21\u65b9x-sin\u76842\u6b21\u65b9x
\u2234f(x)=cos2x+sin2x=\u221a2sin(2x+\u03c0/4)
\u2235x\u2208[0,\u03c0/2]
\u2234\u03c0/4 \u22642x+\u03c0/4\u2264 5\u03c0/4
f(x)\u7684\u6700\u5c0f\u503c=f(5\u03c0/4)=-\u221a2/2
\u96c6\u5408\u4e3a\ufe5bx\u2160x= -\u03c0/8+k\u03c0/2,k\u2208Z\ufe5c
\u6709\u70b9\u96be\u770b\uff0c\u4e0d\u8fc7\u5e94\u8be5\u80fd\u5e2e\u7684\u5230\u4f60
=(sinx)^4-(sinx)^2+1
=(sinx)^2[(sinx)^2-1]+1
=1-(sinxcosx)^2
=1-(1/4)sin2x
绛旓細sinx鐨勫洓娆℃柟鍔燾osx鐨勫洓娆℃柟绛変簬3/4+(cos4x)/4锛屽彲浠ヨ繍鐢ㄤ笁瑙掑嚱鏁扮殑鎬ц川鍏紡杩涜鍖栫畝銆傝y=(sinx)^4+(cosx)^4锛屽垯鏈;y=(sinx)^4+(cosx)^4=[(sinx)^2+(cosx)^2]^2-2(sinx)^2(cosx)^2=1-(sin2x)^2/2=1-[(1-cos4x)/2]/2=3/4+(cos4x)/4 ...
绛旓細鐢ㄤ笅鍏紡搴旇灏卞ソ 妤间富鎱㈡參鐪嬩竴涓嬪惂锛屽簲璇ヤ細鏈変粈涔堥敊璇紝鎴栬鍙互甯垜鎵惧埌 鏈夐棶棰樻杩庤拷闂紝鏈夐敊璇璋㈡寚鏁 娌℃湁闂锛岄敊璇殑璇濋噰绾充竴涓嬪惂
绛旓細鍏跺疄(sin(x))^n鐨勫懆鏈熸槸2*pi/n ;(cos(x))^n鐨勫懆鏈熸槸2*pi/n;鏄剧劧锛氾紙sin(x))^4 鐨勫懆鏈熸槸2*pi/4=pi/2;(cos(x))^4鐨勫懆鏈熸槸 2*pi/4=pi/2锛涙樉鐒朵袱涓嚱鏁扮浉鍔犵殑璇濆懆鏈熷彇鏈澶х殑閭ｄ釜涓,鍥犱负杩欓噷鐩稿姞鐨勪袱涓懆鏈熷嚱鏁扮殑鍛ㄦ湡鐩哥瓑鎵浠ュ懆鏈熶负锛 pi/2 ...
绛旓細y=(sinx)^4+(cosx)^2 =(sinx)^4-(sinx)^2+1 =(sinx)^2[(sinx)^2-1]+1 =1-(sinxcosx)^2 =1-(1/4)sin2x
绛旓細鍋跺嚱鏁帮紱鏈澶у硷細1锛屾渶灏忓硷細1/2锛涙渶灏忔鍛ㄦ湡锛毾/2
绛旓細y(x) = [sin(x)]^4 + [cos(x)]^4 = {[sin(x)]^2 + [cos(x)]^2}^2 - 2[sin(x)]^2[cos(x)]^2 = 1 - [2sin(x)cos(x)]^2/2 = 1 - [sin(2x)]^2/2,1 >= y(x) >= 1 - 1/2 = 1/2 y(0) = 1, y(PI/4) = 1/2,鍑芥暟y=sinx鐨4娆℃柟+cosx鐨4...
绛旓細sinX鐨4娆℃柟鏄伓鍑芥暟锛屽叾鏈灏忔鍛ㄦ湡鐨勏/2锛宑osX鐨2娆℃柟涔熸槸鍋跺嚱鏁帮紝鍏舵渶灏忔鍛ㄦ湡涔熸槸蟺/2锛屼袱涓伓鍑芥暟鐩稿姞杩樻槸鍋跺嚱鏁帮紝涓斿叾鍛ㄦ湡涓鏍凤紝鏁咃紝y=sinX鐨4娆℃柟+cosX鐨2娆℃柟鐨勬渶灏忔鍛ㄦ湡鏄/2
绛旓細y=(sinx)^4+(cosx)^4 =[(sinx)^2+(cosx)^2]^2-2(sinxcosx)^2 =1-1/2*(sin2x)^2 =1-1/4*2(sin2x)^2 =1-1/4(1-cos4x)=3/4+1/4*cos4x 褰4x鈭圼2k蟺-蟺锛2k蟺]锛寈鈭圼(2k-1)蟺/4锛宬蟺/2]鏃讹紝鍗曡皟閫掑
绛旓細瑙ｏ細y=sin^4 x + cos²x =sin^4 x - sin²x +1 =(sin²x -1/2)² +3/4 =[(1-cos2x)/2 -1/2]² +3/4 =cos²2x /4 +3/4 =[(1+cos4x)/2] /4 +3/4 = cos4x / 8 +7/8 鍒欐渶灏忔鍛ㄦ湡涓 2蟺/4=蟺/2.~~~绁濅綘瀛︿範杩涙锛...
绛旓細涓嶇煡閬撲綘鑳界湅娓呭浘鐗囧悧锛熺湅涓嶄翰鐐瑰嚮涓涓嬭繖鏄笉鐢ㄩ珮鏁扮殑鏈绠鍗曠殑鏂规硶浜嗐傝鐢ㄥ埌鍩烘湰涓嶇瓑寮忋傚阀濡欏噾sin^2(x)+cos^2(x)=1鏉ュ仛銆傚綋鐒朵綘涔熷彲浠ョ敤姹傚鏉ュ仛锛屽鏋滃杩囩殑璇濓紝閭ｄ釜鎯冲嚭鏉ュ揩锛岃В涔熷揩锛屼絾鎶鏈惈閲忓氨鐩稿浣庝簡浜涖傛垜鍙仛浜嗘渶澶у硷紝濡傛灉瑕佹眰鏈灏忓硷紝閬撶悊鏄竴鏍风殑锛屽彧鏄瓟妗堝姞涓礋鍙枫
