sincos

首先将cos9分之13派转化为分之4派，然后用公式解决，具体的公式是cos(a+b)=cosacosb-cosacosb,然后把a=派，b=4派除以9分别带进去就可以了。

解：∵sinB=sin（A+C）=sinAcosC+cosAsinC=cosAsinC，∴sinAcosC=0，∵sinA＞0，∴cosC=0，∴C=90°，则S△ABC=1/2ab=6，ab=12，向量AB·AC=cbcosA=9，a²=c²+b²-2cbcosA=c²+b²-18，c²-a²+b²=18，2b²=18，b=3，a...

一个是x2+y2=9x,一个是x2+y2=16y，圆心为（4.5，0）和（0，8），距离√337/2

∵sinα=4/√41,cosβ=9/√82,α和β是锐角，∴cosα=√(1-16/41)=5/√41,sinβ=√(1-81/82)=1/√82,∴cos(α+β)=cosαcosβ-sinαsinβ=(5/√41)×(9/√82)-(4/√41)×(1/√82)=√2/2，∴α+β=π/4。

8.56/75 9.-4/5 10.等腰 11.π/2 12.（π/3，4π/3）13.-4/5

cos13π/9=cos(2π-5π/9)=cos5π/9 =cos(π/2+π/18）=-sinπ/18.

5. sin^4 θ+cos^4 θ=（sin²θ+cos²θ）²－2cos²θsin²θ=1-sin²2θ/2 cos2θ=√2/3 sin²2θ=1-2/9=7/9 sin^4 θ+cos^4 θ=1-7/18=11/18 6.sinθ/2+cosθ/2=2√3/3平方 1+2sinθ/2cosθ/2=4/3 sinθ...

+sinx=cos(9/(2π))cosx-(sin(9/(2π))-1)sinx =(-√(cos(9/(2π)))^2+(sin(9/(2π))-1)^2)cos(x-c)tanc=(sin(9/(2π))-1)/cos(9/(2π))ymin=-√(cos(9/(2π)))^2+(sin(9/(2π))-1)^2 ymax=√(cos(9/(2π)))^2+(sin(9/(2π))-1)^2 ...

如图

sin(9兀/4)+cos(2兀/3)+tan(5兀/4)=sin(2π+π/4)+cos(π-π/3)+tan(π+π/4)=sinπ/4-cosπ/3+tanπ/4 =√2/2-1/2+1 =√2/2+1/2