1 sin163+sin223+sin253+sin313=?2 函数y=sin^4(x)+cos^2(x)的最小正周期?3 求值：1）分子是sin7+cos15sin8 分母是cos7+sin15sin8 2）分子是2cos10-sin20 分母是cos20以上角度省略了"°" 要有思路和过程

sin163+sin223+sin253+sin313
=sin163+sin223+sin(163+90)+sin(223+90)
=sin163+sin223+cos163+cos223
=2sin[(163+223)/2]cos[(163-223)/2]+2cos[(163+223)/2]cos[(163-223)/2]
=2sin193cos30+2cos193cos30
=2cos30*(sin193+cos193)
=2cos30*根号2*sin(193+45)
=根号6*sin238
=-根号6*sin58
y=sin^4(x)+cos^2(x)
=sin^2(x)*(1-cos^2(x))+cos^2(x)
=sin^2(x)+cos^2(x)-sin^2(x)*cos^2(x)
=1-1/4*sin^2(2x)
=1-1/8*(1-cos(4x))
所以周期为Pi/2
分子=sin7+sin(15+8)-sin(15-8)=sin23
分母=cos7-cos(15+8)+cos(15-8)=2cos7-cos23
我怀疑题目中的cos7是不是应该改为-cos7
第二小题，我觉得已经是很简的形式了

1.
sin163度sin223度+sin253度sin313度
=sin17度sin(-43度)+sin(-73度)sin(-47度)
=-sin17度sin43度+sin73度sin47度
=-sin17度sin43度+cos17度cos43度
=cos(17度+43度)
=cos60度
=1/2
2.
y=sin^4(x)+cos^2(x)
=[sin^2(x)]^2+[1-sin^2(x)]
={[sin^2(x)]^2-sin^2(x)}+1
={sin^2(x)[sin^2(x)-1]}+1
={sin^2(x)[-cos^2(x)]}+1
=-(sinxcosx)^2+1
=-(1/4)(2sinxcosx)^2+1
=-(1/4)(sin2x)^2+1
=-(1/4)*[(1-cos4x)/2]+1
=(1/8)cos4x+(7/8)
则：
最小正周期=2π/4=π/2
3.
(1)
[sin7+cos15sin8]/[cos7-sin15sin8]
=[sin(15-8)+cos15sin8]/[cos(15-8)-sin15sin8]
=[(sin15cos8-sin8cos15)+cos15sin8]/[(cos15cos8+sin15sin8)-sin15sin8]
=[sin15cos8]/[cos15cos8]
=[sin15]/[cos15]
=2-√3
(2)
(2cos10-sin20)/cos20
=(2cos10-cos70)/cos20
=[cos10+(cos10-cos70)]/cos20
=[cos10+2sin40*sin30]/cos20
=[cos10+2*(1/2)*sin40]/cos20
=[cos10+sin40]/cos20
=[cos10+cos50]/cos20
=[2cos30*cos20]/cos20
=2cos30
=根号3