given-f-x-2-1-sin-x-2-1-cos-x-find-masimum-value-of-function-f-x-2-

Question Number 85130 by john santu last updated on 19/Mar/20

given f (x) = (\sqrt{2} + 1) \sin x + (\sqrt{2} - 1) \cos x

find masimum value of function

{[f (x)]}^{2}

Answered by Rio Michael last updated on 19/Mar/20

f (x) = a \sin x + bcos x can be expressed in

the form Rsin (x + α) where R > 0 and 0^{°} < α < 90 °

if R is the maximum value of f (x) then R^{2} will be the

maximum value of {[f (x)]}^{2}

\Rightarrow (\sqrt{2} + 1) \sin x + (\sqrt{2} - 1) \cos x = Rsin x \cos α + R \cos x \sin α

\Rightarrow R \cos α = (\sqrt{2} + 1) \dots \dots . (i)

R \sin α = (\sqrt{2} - 1) \dots \dots \dots . (i i)

R^{2} = {(\sqrt{2} + 1)}^{2} + {(\sqrt{2} - 1)}^{2}

R^{2} = 2 + 2 \sqrt{2} + 1 + 2 - 2 \sqrt{2} + 1 = 4 + 2 = 6

\Rightarrow maximum value of {[f (x)]}^{2} = 6

Commented by john santu last updated on 19/Mar/20

thank you

Commented by Rio Michael last updated on 19/Mar/20

welcome sir

Answered by jagoll last updated on 19/Mar/20

if my method

f (x) = k \cos (x - θ)

with k = \sqrt{{(\sqrt{2} - 1)}^{2} + {(\sqrt{2} + 1)}^{2}}

so \max value of {[f (x)]}^{2} equal to

k = 3 - 2 \sqrt{2} + 3 + 2 \sqrt{2} = 6

Commented by john santu last updated on 19/Mar/20

good . thank [you