Question

高一数学题.....急....在线等，谢谢大家了

1、求证:2sin四次方x+¾sin²2x+5cos四次方x-cos3xcosx=2+2cos²x
2、设a、β为锐角，且a=(sina,-cosa),b=(-cosβ,sinβ),a+b=(√6/6，√2/2)试求a*b和cos(a+β)的值
3、函数y=sinxcosx+sinx+cosx求最大值时x的值为
4、若0＜a＜π，sina+cosa=1/2,z,则cos2a的值为

拜托了，哥哥姐姐们，要过程....

Answer 1

2sin^4x+3/4sin^2(2x)+5cos^4x-cos3xcosx=2+2cos²x
等式左边=2sin^4x+3sin^2xcos^2x+5cos^4x-(cos2xcosx-sin2xsinx)cosx
=sin^2x(2sin^2x+3cos^2x)+5cos^4x-(2cos^2x-1-2sin^2x)cos^2x
=2sin^2x+sin^2xcos^2x+5cos^4x-cos^2x+4sin^2xcos^2x
=2sin^2x+5sin^2xcos^2x+5cos^4x-cos^2x
=2sin^2x+5cos^2x-cos^2x
=2+2cos^2x
所以，2sin^4x+3/4sin^2(2x)+5cos^4x-cos3xcosx=2+2cos²x
2.这个写法没见过，姑且先按照我的想法算吧
设a、β为锐角，且a=(sina,-cosa),b=(-cosβ,sinβ),a+b=(√6/6，√2/2)
则有：sina-cosβ=√6/6,sinβ-cosa=√2/2
(a+b)^2=6/36+2/4=2/3,a^2=1,b^2=1,2ab=-4/3,ab=-2/3
利用cos^2β+sin^2β=1化简
得到sin(a-60°)=√6/6,同样可以求出sin(β-30°)=√6/6
a-60=β-30,a=30+β,cos(a-β)=cos30°=√3/2
(sina-cosβ)*(sinβ-cosa)=√12/12=√3/6
sinasinβ-sinacosa-cosβsinβ+cosacosβ=√3/6
sin(a+β)-(sin2a+sin2β)/2=√3/6
因为sin2a+sin2β=2sin(a+β)cos(a-β)
sin(a+β）-√3sin(a+β)=√3/6
sin(a+β）=-√3/[6(√3-1)]=-(√3+3)/12
cos(a+β)=√[1-sin^2(a+β)]
3。y=0.5sin2x+√2sin(45°+x)
x=2kπ+π/4时，Ymax=(1+2√2)/2
4. sina+cosa=√2sin(π/4+a)=1/2,sin(π/4+a)=√2/4
135<45°+a<150°
90°<a<105°，180°<2a<210°
2sinacosa+1=1/4
sin2a=-3/4
cos2a=-√(1-sin^2a)=-√7/4

Answer 2

题多分少！