cosx乘以sinx可以展开吗 展开等于多少 求极限时，如果sinx乘以cosx ，能把sinx泰勒展开然...
sinx\u4e58\u4ee5cosx \u600e\u4e48\u6c42\u51fasinxcosx=1/2(2sinxcosx)=1/2sin2x
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楼主您好：由二倍角公式：sin(2x)=2cosxsinx; 所以cosxsinx=1/2 *sin(2x)可以变成1/2sin2x
绛旓細灏嗗嚱鏁sinxcosx灞曞紑涓簒鐨勫箓绾ф暟鏃讹紝x^3鐨勭郴鏁版槸-2/3 鏍规嵁棰樻剰璁＄畻濡備笅锛歠(x)=sinxcosx =0.5sin2x =0.5[2x-(2x)³/3!+(2x)^5/5!-...=x-2x³/3+2x^5/*58-...鏁厁³鐨勭郴鏁颁负-2/3 鍜岃鍏紡锛歴in ( 伪 卤 尾 ) = sin伪 路 cos尾 卤 cos伪 路 sin尾 ...
绛旓細sinx涔榗osx鑳鐢ㄧ瓑浠锋棤绌峰皬銆俢os(sinx)= cos[x -(1/6)x^3 +o(x^4)]= 1 - (1/2)[x -(1/6)x^3]^2 + (1/24)[x -(1/6)x^3]^4 +o(x^4)= 1 - (1/2)[x^2 -(1/3)x^4 +o(x^4)]+ (1/24)[x^4+o(x^4)] +o(x^4)=1- (1/2)x^2 + ( 1/6 ...
绛旓細sinx*cosx=1/2sin2A鍙敤浜庝笁瑙掑嚱鏁板叕寮忔眰寰椼2sinAcosA=sin2A sinAcosA=1/2sin2A
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绛旓細鏂规硶濡備笅锛岃浣滃弬鑰冿細鑻ユ湁甯姪锛岃閲囩撼銆
绛旓細涓嶆槸锛sinx闄や互cosx绛変簬tanx锛屾垜鐖变綘鍛
绛旓細鍦▁=0澶勬槸杩欐牱锛歴in(x) = x - x ^ 3 / 3! + x ^ 5 / 5! - x ^ 7 / 7! + x ^ 9 / 9! - x ^ 11 / 11! +...+(-1)^k*x^(2k-1)/(2k-1)!+o(x^(2k-1))cos(x)=1-x^2/2!+x^4/4!+...+(-1)^kx^2k/(2k)!+o(x^(2k))鍦▁涓嶇瓑浜0澶勫氨涓嶆槸...
绛旓細鏂规硶涓锛氫緷涔樻硶鍘熷垯 (f(x)*g(x))'=f'(x)*g(x)+f(x)*g'(x)鐢眣=sinx *cosx 鏈墆'=cosx * cosx + sinx * (-sinx)=cos2x 鏂规硶浜岋細鍏堝彉涓涓嬪舰鍐嶆眰瀵 y=sinx * cosx = sin2x /2 y'=cos2x
绛旓細y=sinxcosx=(1/2)sin2x=鈭(1/2)* (-1)^(n-1) *(2x)^(2n-1)/(2n-1)!
