x^2+2*x*sin(x-pi/12)+2*cos(x-pi/12)=(pi-pi/12)^2-2
- 教育綜合
- 2023-04-07 07:56:55
函數(shù)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數(shù)字與字符串問題
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,有效數(shù)字為11位 y=vpa(x/(2*pi))%%計(jì)算y的值,有效位為11位sin^2 (x+π/12)+sin^2 (x-π/12)
sin^2 (x+π/12)-sin^2 (x-π/12) 學(xué)過二倍角公式吧! 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 =[根號(3)sin2x]/2求解高一數(shù)學(xué)題:y=cos2(x-π/12)+sin2(x+π/12),化簡。(函數(shù)式中“2”是平方的意思,嘻嘻打不出來)
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,我這里的2就是數(shù)字2,不是平方(sin^2 x-x^2*cos^2 x)/(x^2*sin^2 x) 關(guān)于x趨近于0的極限 ,用洛必達(dá)法則
我用了洛必達(dá)法則計(jì)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),洛必達(dá)法則 =展開全文閱讀