已知cos(-π 6)+sin=4根号3 5

cos(α-π/6)+sinα=cosα*√3/2+1/2*sinα+sinα=cosα*√3/2+3/2*sinα=√3(sinπ/6cosα+cosπ/6sinα)=√3sin(π/6+α)=4/

再问：再问：在你答题的时候我蛋疼做了一遍，结果好像不一样……再问：不过还是辛苦施主了

(1)f(x)＝sin(2x＋π/6)＋sin(2x－π/6)＋2cos²x＝sin2xcosπ/6＋cos2xsinπ/6＋sin2xcosπ/6－cos2xsinπ/6＋2cos&sup

1.f(x)=sin(2x+6/π)+2sin^2x=(√3/2)sin2x+(1/2)cos2x+1-cos2x=(√3/2)sin2x-(1/2)cos2x+1=sin(2x-π/6）+1所以f(

多一个sin(2x+π/6)吧!f(x)=sin(2x+π/6)+2cos²x=sin2xcosπ/6+cos2xsinπ/6+1+cos2x=√3/2sin2x+3/2cos2x+1=√3

cos(α-π/6)+sinα=4√3/5cos(α-π/6)+cos(π/2-α)=4√3/52cos{[(α-π/6)+(π/2-α)]/2}*cos{[(α-π/6)-(π/2-α)]/2}=4

解应为（sinα+cosα）/(sinα-cosα)=2两边平方得（sin²α+cos²α+2sinαcosα）/(sin²α+cos²α-2sinαcosα)

sinα-cosα=1/5,则有（sinα-cosα）^2=1/25即(sinα)^2+(cosα)^2-2sinαcosα=1/25sin2α=1-1/25=24/25cos2α=√(1-(sin2

sinα+cosα=1==>sin²x+2sinxcosx+cos²x=1sinxcosx=0所以sinx=0或cosx=0,cosx=±1或sinx==±1(sinα)^6+(c

sinα-cosα=-根号5/5π

后面化出来是2-3sinacosa-(cosa)^2=2-cosa(3sina-cosa),前面哪个式子同时除以cosa可以得出tana的一个等式,求出来tana=1,证明a=π/4+kπ,那么代入前

6sin²α+5sinαcosα-4cos²α=0（2sina-cosa)(3sina+4cosa)=02sina-cosa=0或3sina+4cosa=02sina=cosata

6sin²α+sinαcosα-2cos²α=0,α∈（π/2,π）,同时除以cos²α得6tan²α+tanα-2=0,故tanα=1/2或-2/3α∈（π/

2sin(a/2)=cos(a/2）tan(a/2)=1/2sinα=2tan(α/2)/[1+tan^2(α/2)]=1/（1+1/4）=4/5cosα=[1-tan^2(α/2)]/[1+tan^

f(x)=cos(π/6-2x)+cos(2x+π/6)+sin(2x+π/3)-sin(π/3-2x)=sin(π/2-(π/6-2x))+cos(2x+π/6)+sin(2x+π/3)-cos(π

sin(2α+π/3)=3*(根号3)-4/10或4*(根号3)+3/106sin^2α+sinαcosα-cos^2α=0(2sinα+cosα)(3sinα-cosα)=02sinα+cosα=0

sin(x-π/3)-cos(x+π/6)+√3cosx=sinxcosπ/3-cosxsinπ/3-cosxcosπ/6+sinxsinπ/6+√3cosx=1/2*sinx-√3/2*cosx-√

根据题意可知：sinx>0cosx

[cos^4(π/3+α)]-[cos(π/6-α)]^2=[cos(π/3+α)]^4-[sin(π/3+α)]^4=[cos(π/3+α)]^2-[sin(π/3+α)]^2=cos(2π/3+2

解:⑴f(x)=-1/2+sin(π/6-2x)+cos(2x-π/3)+(cosx)^2=-1/2+sinπ/6cos2x-sin2xcosπ/6+cos2xcosπ/3+sin2xsinπ/3+(