tanx=sin(x 二分之π),则sinx=

f(x)=√2cos(x/2)-(a-1)sin(x/2)f(-x)=√2cos(x/2)+(a-1)sin(x/2)f(x)=f(-x)a-1=0a=1f(x)=√2cos(x/2)T=2π/ω=2

a=(cos(3x/2),sin(3x/2))b=(cos(x/2),-sin(x/2))因此,a·b=cos(3x/2)cos(x/2)-sin(3x/2)sin(x/2)=cos(3x/2+x/2

属于0*∞型,变形后用罗比塔法则：lim(x-->π/2)（sinx-1)tanx=lim(x-->π/2)（sinx-1)/cotx=lim(x-->π/2)cosx/[-csc^2(x)]=lim

先用tanx=sinx/cosx、倍角公式、诱导公式化简原函数：f（x）=sin²x+sinxcosx-sin[2（x+π/4）]=（1-cos2x）/2+1/2sin2x-sin（2x+π

利用tanx=sinx/cosx的定义,左边的一定可以化为sinx和cosx的式子同理右边的可以分解为sinx和cosx的式子二者一定相等

tanx=(2tan二分之x)/(1-tan二分之x的平方),答案是负三分之四.由上面的式子,而且tanx=-4/3,tan(四分之派)=1,所以答案为负七分之一.

设x-π/2=tlim(x->π/2)ln[x-π/2]/tanx=lim(t->0)lnt/tan(t+π/2)=lim(t->0)lnt/-cott(无穷/无穷型,用洛必达)=lim1/t/-(-

sin二分之π,cos二分之π,tan二分之π不是等于1,0,∅

φ=二分之π则f(x)=cosx,是偶函数所以是充分而f(x)是偶函数则φ=二分之π+kπ所以不是必要所以是充分非必要条件

tanx/tanx-1=-1tanx=-tanx+1tanx=1/2sinx/cosx=1/22sinx=cosx平方4sin^2x=4(1-cos^2x)=4-4cos^2x=cos^2xcos^2

f(x)=cos(3x/2)cos(x/2)-sin(3x/2)sin(x/2)-2sinxcosx=cos(3x/2+x/2)-2sinxcosx=cos2x-sin2x=√2(√2/2*cos2x

f(X)=2sinxcosx+cos2xf(α/2)=√2/22sinα/2cosα/2+cosα=√2/2sinα+cosα=√2/2cosα=√2/2-sinα1-sin^2α=1/2-√2sin

(sinx)^2tanx=[1-(cosx)^2]tanx=tanx-(cosx)^2tanx=tanx-(cosx)^2*sinx/cosx=tanx-sinxcosx(cosx)^2cotx=[1

f(x)=(1+1/tanx)*(sinx)^2-2sin(x+π/2)sin(x-π/4)=(1+cosx/sinx)*(sinx)^2+2sin(x+π/4)cos[(x-π/4)+π/2]=(s

1.证明:角BAC为直角,即,证明:向量AB*向量AC=0,即可,向量AB*向量AC=(1+tanx)*sin(x-π/4)+(1-tanx)*sin(x+π/4)=[sin(x-π/4)+sin(x

17π/12＜x＜7π/4,得5π/3＜x+π/4＜2πcos(x-π/4)=cos[(x+π/4)-π/2]=sin(x+π/4)=-√[1-sin²(x+π/4)]=-√[1-(3/5)

请个给我留言,

不等.∵sin(½∏－A)＝cosA.

(1).向量a•向量b=cos(3x/2)cos(x/2)-sin(3x/2)sin(x/2)=cos2x|向量a+b|=[cos(3x/2)+cos(x/2)]^2+[sin(3x/2)