cossinx-cosx用三角公式

蒋通屈3793y=cosx(sinx - cosx)的周期最值 -
喻钧品19224191850 ______ y=cosx(sinx-cosx) = (1/2)sin2x - (1/2)(cos2x+1) = (√2/2)sin(2x-π/4) -1/2 周期=π 最值 max = √2/2 -1/2 min =-√2/2 -1/2

蒋通屈3793以知函数F(x)=2cos(sinx - cosx)+1求F(x)最小正周期及增区间 -
喻钧品19224191850 ______ F(x)=2cos(sinx-cosx)+1=2cosxsinx+2cos^2x+1=sin2x+cos2x+2=√2sin(2x+π/4)+2 最小正周期:2π/2=π2x+π/4,单调递增区间在[2Kπ-π/2,2Kπ+π/2] x单调递增区间在[Kπ-3π/8,Kπ+π/8]

蒋通屈3793一道高中数学题已知函数F(x)=2cos(sinx - cosx)+
喻钧品19224191850 ______ F(x)=2cos(sinx-cosx)+1=F(x)=2cos(√2cos(x+π/4))+1=2cos|(√2cos(x+π/4))|+1 √2cos(x+π/4)的最小正周期 为2π , |(√2cos(x+π/4))|的最小正周期 为π , 函数F(x)的最小正周期 为π . √2cos(x+π/4)在区间[π/8,π/2]上是单调递减,在区间(π/2,3π/4]上是单调递增的. 因此F(x)在区间[π/8,π/2]上是单调递增,在区间(π/2,3π/4]上是单调递减的. 又因为F(π/8)=2cos(√2cos3π/8)+1,F(π/2)=2cos1+1,F(3π/4)=2cos√2+1. F(π/8)全部

蒋通屈3793sinx - cosx=1/2,则sinx的平方 - cos的平方= -
喻钧品19224191850 ______[答案] (sinx-cosx)^2=1-2sinxcosx=1/4 2sinxcosx=3/4 (sinx+cosx)^2=1+2sinxcosx=1+3/4=7/4 sin+cosx=正负根号7/2 sin^2-cos^2=(sinx+cosx)(sinx-cos)=1/2*(根号7/2)=正负(根号7/4)

蒋通屈3793sinx - cosx=1/3 则sinxcosx= ? -
喻钧品19224191850 ______ (sinx-cosx)²=sin²x+cos²x-2sinxcosx=1/9 ∵sin²x+cos²x=1 所以sinxconx=(1-1/9)/2=4/9

蒋通屈3793化简2cosx(sinx - cosx)+1 -
喻钧品19224191850 ______ f(x)=(1 sinx cosx)[sin(x/2)-cos(x/2)]/√(2 2cosx) =[sin²(x/2) cos²(x/2) sinx cosx][sin(x/2)-cos(x/2)]/[√2(1 cosx)] =[sin²(x/2) cos²(x/2) 2sin(x/2)*cos(x/2) cos²(x/2)-sin²(x/2)][sin(x/2)-cos(x/2)]/[√4*(1 cosx)/2] ={[sin(x/2) cos(x/2)]² [cos(x/2) sin(x/2)]...

蒋通屈37933sinx - cosx怎么化简 -
喻钧品19224191850 ______ 解:原式=√3sinx-cosx=2(sinxcosπ/6-cosxsinπ/6)=2sin(x-π/6)

蒋通屈3793sinxcosx等于八分之一,那么cos - sin等于 -
喻钧品19224191850 ______ ∵(sinx-cosx)²=[-(cosx-sinx)]²=sin²x-2sinxcosx+cos²x=1-2sinxcosx=1-1/4=3/4, ∴sinx-cosx=-(cosx-sinx)=±√3/2,cosx-sinx=±√3/2

蒋通屈3793已知sinx - cosx=1/2如何sin^3x+cos^3x
喻钧品19224191850 ______ 因为sinx-cosx=1/2,则(sinx-cosx)^2=1/4;sin^2+cos^2-2sinxcosx=1-2sinxcosx=1/4得sinxcosx=3/8计算(sinx+cosx)^2=sin^2+cos^2+2sinxcosx=1+2*(3/8)=7/4所以(sinx+cosx)=√(7/4)=√7/2根据a^3+b^3=(a+b)(a^2-ab+b^2) ,sin^2x+cos^2x=1得...

蒋通屈3793已知2sinx= - cosx,则1+2sinxcosx/(sin^x - cos^x)的值是____. -
喻钧品19224191850 ______[答案] 原式=sin^x+cos^x+2sinxcosx/(sin^x-cos^x) =(sinx+cosx)^2/(sinx+cosx)(sinx-cosx) =(sinx+cosx)/(sinx-cosx) 2sinx=-cosx,带入 原式=-1/3