根号下1 2sin(π-2)

sin(a+π/3)+sina=(-4/5)*根号3展开左边,即√3sin(a+π/6)=(-4/5)*根号3∴sin(a+π/6）=-4/5∵-π/2再问：麻烦下左边的式子怎么拆开请帮下。。怎么得到

半角公式化简3π/4

原式=√[1-(-sin2)*(-cos2)]=√(1-sin2*cos2)=√(1-1/2sin4)==√2/2*√(2-sin4)应该化简不了了吧?你题目有没有打错啊?

√[2+2sin(2π-θ)-cos^2(π+θ)]=√[2-2sinθ-cos^2θ]=√[2-2sinθ-(1-sin^2θ)]=√(sin^2θ-2sinθ+1)=√(sinθ-1)^2=1-s

由题意可得sin(2x+π/6)

√[1-2sin(π+4)cos(π+4)]√[sin²(π+4)-2sin(π+4)cos(π+4)+cos²(π+4)]=√{[sin(π+4)-cos(π+4)]²

利用三角函数诱导公式加一项,再分子有理化,过程如下：lim(n→无穷大)sin[根号下(n^2+1)]*π=-lim(n→无穷大)sin{[根号下(n^2+1)]-n}*π=-lim(n→无穷大)si

1-cos²θ=sin²θsin（2π-θ）=-sinθcos（π+θ）=-cosθ2+2sin（2π-θ）-cos²（π+θ）=2-2sinθ-cos²θ=1

是化简根号下1+sinθ—根号下1—sinθ吧?因为θ∈（0,2分之π）,所以θ/2∈（0,π/4）,∴0

由已知：sin(2π-A)=-sinA=-2^0.5*sin(π-B)=-2^0.5*sinB,3^0.5*cos(2π-A)=3^0.5*cosA=-2^0.5*cos(π+B)=2^0.5*cos

根号下(1-cos^2α)+根号下(1-sin^2α)=根号下(sin^2α)+根号下(cos^2α)=|sina|+|cosa|=sinα-cosα.1.a∈[0,π/2].原式化为：sina+co

2sin(x+π/4）+1≥0sin(x+π/4)≥-1/22kπ-π/6≤x+π/4≤2kπ+7π/62kπ-5π/12≤x≤2kπ+11π/12[2kπ-5π/12,2kπ+11π/12],k∈Z

[√(1－2sin2cos2)]－√(1－sin²2)=√[sin2－cos2]²－√(cos²2)=|sin2－cos2|－|cos2|=(sin2－cos2)＋cos

sin(x+π/3)=sinxcos(π/3)+cosxsin(π/3)=1/2sinx+√3/2cosx2sin(x-π/3)=2sinxcos(π/3)-2cosxsin(π/3)=sinx+√3

(1-sinα*sinβ)²-(cosα)²(cosβ)²=(1-sinα*sinβ+cosα*cosβ)*(1-sinα*sinβ-cosα*cosβ)=[1+cos(

1-sin^2（160°）=cos^2(160°)=cos^2(20°）

因为cos2α=sin(π/2-2α)=sin2(π/4-α)=2sin(π/4-α)cos(π/4-α)所以(cos2α)/sin(α-π/4)=-(cos2α)/sin(π/4-α)=-2cos(

sin(π-3)=sin3cos(π+3)=-cos31+2sin(π-3)cos(π+3)=1-2sin3cos3=(sin3)^2-2sin3cos3+(cos3)^2=(sin3-cos3)^2

原式=1+2sin(3)cos(3)=1+sin6奇变偶不变,符号看象限　sin（π+α）=-sinα　　cos（π+α）=-cosα　　tan（π+α）=tanα　　cot（π+α）=cotα