Question

what is the maximum value of 19-cos^2x​

Answer 1

Answer:

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Let y = sinx +cos2x

dy/dx = cosx -2sin2x = cosx -4sinxcosx

=cosx(1–4sinx)

At the points of extremum dy/dx= 0

cosx(1–4sinx) =0 => cos x = 0 or 1–4sinx =0

=> cosx=0 or sinx = 1/4 => x =π/2

or x = sin^(-1) 1/4

Now d^2y/dx^2= cosx.(-4cosx)+(1–4sinx)(-sinx)

=4sin^2 x - 4 cos^2 x-sinx

When x= π/2 , d^2y/dx^2 = 3 > 0. So y is minimum at x = π/2 , and the minimum value of y = 0

When sin x = 1/4 , d^2y/dx^2

=4×(1/16) - 4(1 - 1/16) - 1/4= -15/4 < 0 => y is maximum at x = sin^(-1) (1/4)and the maximum value of y =sin x + cos2x = sinx + 1–2 sin^2 x

=(1/4)+1 - 2×(1/16)= (1/8) + 1 = 1 1/8.

Answer 2

Answer:19

Step-by-step explanation:

X=90

Cos90=0

Cos^2 90=0