x^2+2*x*sin(x-pi/12)+2*cos(x-pi/12)=(pi-pi/12)^2-2

函数f(x)=sin^2(x+pi/12)+cos^2(x-pi/12)的最大值

f(x)=sin²(x+π/12)+cos²(x-π/12) =1-cos²(x+π/12)+cos²(x-π/12) =1+[cos(x-π/12)+cos(x+π/12)][cos(x-π/12)-cos(x+π/12)] =1+(cosx cosπ/12 + sinx sinπ/12 + cosx cosπ/12 -sinx sinπ/12)(cosx cosπ/12 + sinx sinπ/12 - cosx cosπ/12 + sinx sinπ/12) =1 + 2 cosx cosπ/12 × 2 sinx sinπ/12 =1 + 2sinxcosx×2sinπ/12cosπ/

matlab数字与字符串问题

beta=0.2094; equ=['x^2+2*x*sin(x' ,num2str(-beta,'%+f'), ')+2*cos(x',num2str(-beta,'%+f'),')-(pi',num2str(-beta,'%+f'),')^2+2']; x=solve(equ) digits(11);%%%控制精度，有效数字为11位 y=vpa(x/(2*pi))%%计算y的值，有效位为11位

sin^2 (x+π/12)+sin^2 (x-π/12)

sin^2 (x+π/12)-sin^2 (x-π/12) 学过二倍角公式吧！ cos2A=1-2sin^2 (A) 利用此公式, sin^2 (x+π/12)-sin^2 (x-π/12) =[1-cos(2x+π/6)]/2-[1-cos(2x-π/6)]/2 =[cos(2x-π/6)-cos(2x+π/6)]/2 =[(cos2xcosπ/6+sin2xsinπ/6)-(cos2xcosπ/6-sin2xsinπ/6)]/2 =[根号(3)sin2x]/2

求解高一数学题：y=cos2(x-π/12)+sin2(x+π/12)，化简。（函数式中“2”是平方的意思，嘻嘻打不出来）

y=[1+cos2(x-π/12)]/2+[1-cos2(x+π/12)]/2 =[cos(2x-π/6)-cos(2x+π/6)]/2+1 =(cosxcosπ/6+sinxsinπ/6-cos2xcosπ/6+sin2xsinπ/6)/2+1 =sin2xsinπ/6+1 =(sin2x)/2+1 注意，我这里的2就是数字2，不是平方

(sin^2 x-x^2*cos^2 x)/(x^2*sin^2 x) 关于x趋近于0的极限 ，用洛必达法则

我用了洛必达法则计算: lim(x→0) (sin²x-x²cos²x)/(x²sin²x) =lim(x→0) (sinx-xcosx)(sinx+xcosx)/(x²sin²x) =lim(x→0) (sinx+xcosx)x*(sinx-xcosx)/x³*(x/sinx)² =lim(x→0) (sinx/x+cosx)*lim(x→0) (x/sinx)²*lim(x→0) (sinx-xcosx)/x³ =(1+1)*1*1/3 =2/3 其中: lim(x→0) (sinx-xcosx)/x³ =lim(x→0) [cosx-(cosx-xsinx)]/(3x²),洛必达法则 =