函数f(x)=asinx+btanx+1,满足f(5)=7,则f(-5)=?f(cosx)=cos2x 则f(sin15°）的值等于?已知三角形ABC的3内角A,B,C成等差数列,且A-C=π/3,求cos^2A+cos^2B+cos^2C的值 （cos^2A就是cosA的平方哈）

1.令F(x)=f(x)-1=asinx+btanx
F(x)为奇函数
F(-5)=-F(5)=-7
所以f(5)=-6
2.cosx=t
f(t)=2t^-1
f(sin15°)=2sin^15°-1=-cos30°=负二分之根三
3.2B=A+C
B=60°A=90° C=30°
cos^2A+cos^2B+cos^2C=1

1)F(x)=f(x)-1=asinx+btanx为奇函数，F(5)=7-1=6, 故F(-5)=-6
f(-5)=-6+1=-5
2)f(sin15°)=f(cos75°)=cos150°=-√3/2
3)3B=π得B=π/3， A+C=2π/3, A-C=π/3, 得A=π/2，C=π/6
cosA=0, cosB=1/2, cosC=√3/2
所求式为1

f(x)-1=asinx+btanx
f(5)-1=asin5+btan5
f(-5)-1=asin(-5)+btan(-5)=-asin5-btan5=-(f(5)-1)
f(-5)-1=-(7-1)=-6
f(-5)=-5
f(sinx)=f(cos(90-x))=cos2(90-x)
f(sin15）=cos2(90-15)=cos150
等差数列,有2B=A+C,3内角和A+B+C=π,加上A-C=π/3
解得A=π/2,B=π/3,C=π/6
cos^2A+cos^2B+cos^2C=0+1/4+3/4=1