Maximum and minimum of $f(x) = 2\sin^2 x + 2\cos^4 x$

Question

I modified the equation into $2\sin^2 x + 2(1 - \sin^2 x)^2$

let $\sin^2x$ as $p$.

so, $f(p) = 2p + 2(1 - p)^2$

max and min at $p = \pm \frac 12$

$f(-\frac 12)$ and $f(\frac 12)$ give the same value is $\frac 32$

also $-1 \leq p \leq 1$ so, $p(-1)$ and $p(1)$ give the same value is $2$

so max is $2$ and min is $\frac 32$

My question is

should we always check and included $\sin x = -1$ or $\sin x = 1$ for trigonometry equation similar to this (lets say theres no certain boundary) ?

at slope $= 0$, there's only $1$ value, so it means on the graph has only 1 peak that is maxima or minima because this function is quadratic equation?

so $\frac 32$ is global minima? and $2$ is global maxima?

Answer 1

You also could linearise the function first, doing some trigonometry: \begin{align} 2\sin^2x+2\cos^4x&=1-\cos 2x+\frac12(2\cos^2 x)^2=1-\cos 2x+\frac12+\cos 2x+\frac12\cos^22x \\ &=\frac32+\frac{1+\cos4 x}4=\frac{7+\cos 4x}4. \end{align} Now it is obvious this expression has a global minimum when $\cos 4x=-1$ and a global maximum when $\cos 4x=1$.

Answer 2

Your idea is good, but you missed one thing: since $p=\sin^2x$, the set of the possible values of $p$ is $[0,1]$, not $[-1,1]$. But, yes, the minimum is $\frac32$ and the maximum is $2$.

Answer 3

$f(x)=2\sin^2+2(1-\sin^2)\cos^2x=$

$2(\sin^2x+\cos^2x)-2\sin^2x\cos^2$;

$f(x)=2-(1/2)\sin^2(2x)$;

Minimum:

at $2x=k2π+π/2, k \in \mathbb{Z};$

$f_{\min}(x)=3/2$;

Maximum:

at $2x=k2π, k \in \mathbb{Z};$

$f_{\max}(x)=2$.

Answer 4

Let $\cos^2x=q$

$$f(q)=2(1-q)+2q^2=2(q-1/2)^2+2-1/2$$

Now $0\le q\le1\implies -1/2\le q-1/2\le1-1/2$

$\implies0\le(q-1/2)^2\le(1/2)^2$

Answer 5

First we may notice that the given function has a period of $\pi$ .Then $f'=\sin{4x}$, with critical values $0,\pi/4, \pi/2, 3\pi/4, \pi$.From these we can conclude all of what you said.