sin2π+x

贾版俊3798函数sin(x+π/4) - sin2x(x∈R)最大值是 -
许孟沈15177882846 ______ sin2x=-cos(2x+π/2)=-cos[2(x+π/4)]=2sin^2 (x+π/4)-1 u= sin(x+π/4),|u|≤1, y=u-(2u^2-1)=-2(u-1/4)^2+9/8 u=1/4,y max=9/8

贾版俊3798'已知sin(x+π/4)= - 5/13,则sin2x的值 -
许孟沈15177882846 ______ sin(π/4+x)=-5/13 sinxcosπ/4+cosxsinπ/4=-5/13 √2/2(sinx+cosx)=-5/13 sinx+cosx = -5√2/13(sinx+cosx)² = (-5√2/13)² sin²x+cos²x+2sinxcosx=50/1691+sin2x=50/169 sin2x = -119/169 如果你认可我的回答,请点击“采纳为满意答案”,祝学习进步!手机提问的朋友在客户端右上角评价点【满意】即可

贾版俊3798函数y= sin2x+2√2cos(π/4+x)+3的最大值是? -
许孟沈15177882846 ______[答案] y' = 2cos2x-2√2sin(x+π/4)=0 y''=-4sin2x - 2√2cos(x+π/4) cos2x=√2sin(x+π/4) cos2x=(sinx+cosx) x=0 y''(0) = -2 因此 最大值:y(0)=4

贾版俊3798已知cos(π/4+x)=3/5,5/4π许孟沈15177882846 ______[答案] cos(π/4+x)=3/5,3/2π<π/4+x<2π, 所以sin(π/4+x)=-4/5, Sin2x=-cos(π/2+2x) =-cos[2(π/4+x)] =-[1-2sin²(π/4+x)]=7/25. sinx+cosx=√2* sin(π/4+x)=-4√2/5, cosx –sinx=√2* cos(π/4+x)=3√2/5, (sin2x+2(sinx)^2)/(1-tanx) =[2sinx(sinx+cosx)]/(1-sinx/cosx)……...

贾版俊3798sin2x+2sin(45°+x)的最小值 -
许孟沈15177882846 ______ sin2X + 2sin(X + 45°)=2sinXcosX + 2(√2/2 sinX + √2/2 cosX)=2sinXcosX + √2(sinX + cosX)=2sinXcosX + √2(sinX + cosX) +1 - 1=2sinXcosX + √2(sinX + cosX) + sin^2X + cos^2X - 1=(sinX + cosX)^2 + √2(sinX + cosX) - 1 设(sinX + ...

贾版俊3798已知cos(π/4+x)= - 3/5,3/4π许孟沈15177882846 ______[答案] cos(π/4+x) =cosπ/4cosx-sinπ/4sinx =(√2/2)(cosx-sinx) =-3/5 cosx-sinx=-3√2/5 (cosx-sinx)²=18/25 1-2sinxcosx=18/25 2sinxcosx=7/25 (sinx+cosx)²=1+2sinxcosx=1+7/25=32/25 又∵3/4π

贾版俊3798已知函数f(x)=cos2x+sin2x,x属于R. 求函数f(x)在区间〔 - π/8,π/2〕的最小值和最大值,并求出取得最值时x的 -
许孟沈15177882846 ______ f(x)=cos2x+sin2x=√2[(sin2x)*(√2/2)+(cos2x)*(√2/2)] =√2[sin2xcos(π/4)+cos2xsin(π/4)] =√2sin(2x+π/4) -π/8≤x≤π/2 ==> 0≤2x+π/4≤5π/4 -√2/2≤sin(2x+π/4)≤1 ==> -1 ≤ f(x) ≤ √2 f(MAX)=√2 当且仅当 2x+π/4=π/2+2kπ 即x=π/8+kπ 时取“=”本题是x=π/8 f(min) =-1 当且仅当 2x+π/4=5π/4+2kπ 即x=π/2+kπ 时取“=”本题是x=π/2

贾版俊3798已知cos(π/4 +x)=3/5,17π/12<x<7π/4,求(sin2x+2sin??x)/1 - tanx的值 -
许孟沈15177882846 ______ 您好:解答如下17π/12

贾版俊3798(SIN2X)^2+SIN2X+COS2X=1求角X -
许孟沈15177882846 ______ sin²2x+sin2x+cos2x=1 sn2x+cos2x=1-sin²2x=cos²2x sin2x=cos²2x-cos2x ---------------------------------------------(1)(sin2x)²=(cos²2x-cos2x)² sin²2x=(cos²2x-cos2x)²1-cos²2x=cos²2x(cos2x-1)² 设:cos²2x=t,则:1-t=t(t-1)² t(t-1)²+(t-...

贾版俊3798化简f(x)=cos^2(x+π/12)+1/2sin2x -
许孟沈15177882846 ______ f(x)=cos^2(x+π/12)+1/2sin2x =1/2*[1+c0s(2x+π/6)]+1/2sin2x =1/2+1/2(cos2xcosπ/6-sin2xsinπ/6)+1/2sin2x =1/2+1/2cos2xcosπ/6-1/4sin2x+1/2sin2x =1/2+1/2cos2xcosπ/6+1/4sin2x =1/2+1/2(cos2xcosπ/6+sin2xsinπ/6) =1/2+1/2cos(2x-π/6) = 1/2[1+cos(2x-π/6)] =cos^2(x-π/12)