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

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

Let x = cosa
Then y & # 178; = 1-cos & # 178; a = Sin & # 178; a
So y = Sina
So 3x-4y = - 4sina + 3cosa
=-√(4²+3²)sin(a-b)
=-5sin(a+b)
Where tanb = 3 / 4
So the maximum is 5