Question

y=（3sinx-3)/(2cosx+10)的最大值和最小值

Answer 1

令: t = tg(x/2), 则: sinx = 2t/(1+t^2), cosx = (1-t^2)/(1+t^2)

y = (3sinx-3)/(2cosx+10) = {3*[2t/(1+t^2)]-3}/{2*[(1-t^2)/(1+t^2)]+10}

= (-3/8)[1 - (4t+1)/(2t^2+3)]

令: A = (4t+1)/(t^2+3)

则: 2At^2 - 4t + (3A-1) = 0

4^2 - 4*(2A)*(3y-1) >= 0

-2/3