Find the range of $f(x)=11\cos^2x+3\sin^2x+6\sin x\cos x+5$

Question

I'm trying to solve this problem.

Find the range of $f(x)=11\cos^2x+3\sin^2x+6\sin x\cos x+5$

I have simplified this problem to $$f(x)= 8\cos^2x+6\sin x\cos x+8$$ and tried working with $g(x)= 8\cos^2x+6\sin x\cos x$.

I factored out the $2\cos x$ and rewrote the other factor as a linear combination of cosine.

It reduces down to $$g(x)=10\cos x\cos\left(x-\tan^{-1}\frac{3}{4}\right)$$

But then I'm stuck here. Please help me. Perhaps there's a different way to approach this?

Answer 1

Rewrite as following

$$\begin{align} f(x) & = 8 + 6 \sin(x ) \cos(x ) + 8\cos^2(x) - 4 + 4 \\ &= 12 + 3 \sin(2x) + 4 \cos(2x) \\ &= 12 + 5 \left(\frac{3}{5} \sin(2x) + \frac{4}{5} \cos(2x)\right) \\ &= 12 + 5 \sin(2x + \arctan{\tfrac{4}{3}}) \end{align}$$

Now its easy since $\sin(...)$ always lies in $[-1,1]$, max/min values are $12 \pm 5$.

So maximum value is $17$ and minimum is $7$

Answer 2

write your function as $$f(x)=(\sin(x)+3\cos(x))^2+7$$

Answer 3

Hint:

Let $y=A\sin^2x+B\cos^2x+C\sin x\cos x+D$

Method$\#1:$ Divide both sides by $\cos^2x$ to form a Quadratic Equation in $\tan x$

As $\tan x$ is real, the discriminant must be $\ge0$

Method$\#2:$ Divide both sides by $\sin^2x$

Answer 4

$$f(x)=8\left(\frac{1+\cos {2x}}{2}\right) +3\sin {2x}+8$$

$$f(x)=4\cos {2x}+3\sin {2x} +12$$

The range of $$4\cos {2x}+3\sin {2x}$$ is $[-5,5]$

Hence maximum and minimum values of expression are $17$ and $7$ respectively.