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

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

• Have you tried finding the maximum and minimum valueof $f$? Jan 6 '18 at 15:44

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 $7$

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

• This is good for minimum, but then for maximum? But answer is good! +1! Jan 6 '18 at 15:59
• You should write $3\cos x + \sin x$ in the form $R\cos(x-\alpha)$, where $R>0$ and $0 < \alpha < \frac{1}{2}\pi$. Jan 6 '18 at 16:12
• you Can write your function as $$f(t)=4\,{\frac {4\,{t}^{4}-3\,{t}^{3}+3\,t+4}{ \left( {t}^{2}+1 \right) ^{2 }}}$$ with $$\tan(\frac{x}{2})=t$$ and use calculus Jan 6 '18 at 16:16
• @Dr.SonnhardGraubner or use that $\sin(x) + 3 \cos(x) \in [-\sqrt{1^2+ 3^2}, \sqrt{1^2+3^2}]$ Jan 7 '18 at 11:51

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

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

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

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

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 $\sin^2x$