Find the range of \(\displaystyle{f{{\left({x}\right)}}}={11}{{\cos}^{{2}}{x}}+{3}{{\sin}^{{2}}{x}}+{6}{\sin{{x}}}{\cos{{x}}}+{5}\) I have simplified

poznateqojh

poznateqojh

Answered question

2022-04-05

Find the range of f(x)=11cos2x+3sin2x+6sinxcosx+5
I have simplified this problem to
f(x)=8cos2x+6sinxcosx+8
and tried working with g(x)=8cos2x+6sinxcosx

Answer & Explanation

Terzago66cl

Terzago66cl

Beginner2022-04-06Added 8 answers

Rewrite as following
f(x)=8+6sin(x)cos(x)+8cos2(x)4+4
=12+3sin(2x)+4cos(2x)
=12+5(35sin(2x)+45cos(2x))
=12+5sin(2x+arctan{43})
Now its easy since sin() always lies in [-1,1], max/min values are 12±5
So maximum value is 17 and minimum is 7
cab65699m

cab65699m

Beginner2022-04-07Added 14 answers

Hint:
Let y=Asin2x+Bcos2x+Csinxcosx+D
Method#1: Divide both sides by cos2x to form a Quadratic Equation in tanx
As tanx is real, the discriminant must be 0
Method#2: Divide both sides by sin2x

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