函数f(x)=asinx+btanx+1,满足f(5)=7,则f(-5)=?

题目

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）

答案

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