If x ^ 2 + y ^ 2 = 1, what is the maximum value of 3x-4y

x^2+y^2=1
set up
x=sina,y=cosa
(sina)^2+(cosa)^2=1
3x-4y=3sina-4cosa
=√ [(4 ^ 2 + 3 ^ 2)] * sin (a + b) (B is any real number)
=5sin(a+b)《5
So the maximum value of 3x-4y is 5

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