Question

if y=10cos^2x-6sinxcosx+sin^2x, then find the greatest & least value of y

Answer 1

y = 10 cos^2x - 6 sinx cosx + 2 sin^2x

y = 9cos^2x - 6 sinx cosx + sin^2x + (cos^2x+sin^2x)

y = 1 + (3cos(x) - sin(x))^2

min y is 1 for 3cos(x) - sin(x) = 0 else for any x , y >= 1

max when 3cos(x) - sin(x) has max absolute value

Let be u = 3cos(x) - sin(x) = sqrt(10)[sin(a)cos(x) - sin(x)cos(a)] where a = arctan(3) then

u = sqrt(10)[sin(a-x)] max absolute value is sqrt(10) then max value for y is y 1+(sqrt(10))^2 = 11

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