# Minimizing $\left ( \sin^2(x) + \frac{1}{\sin^2(x)} \right )^2 + \left ( \cos^2(x) + \frac{1}{\cos^2(x)} \right )^2$

While solving a problem I came across this task, minimizing \begin{align} \left ( \sin^2(x) + \frac{1}{\sin^2(x)} \right )^2 + \left ( \cos^2(x) + \frac{1}{\cos^2(x)} \right )^2. \end{align}

One can easily do it with calculus to show that the minimum value is $$12.5$$. I tried to do it using trigonometric identities and fundamental inequalities (like AM-GM, Cauchy-Schwarz, etc.) but failed. Can someone help me to do it using trig identities and inequalities?

• A method that is not good but works: use $1-\cos^2 x=\sin^2x$ and the thing turns into a rational function of $\sin ^2 x$, which is in $[0,1]$. – Trebor May 1 '19 at 12:18
• – lab bhattacharjee May 1 '19 at 13:06

Knowing the answer it's not that difficult to get it another way. Let $$y = \cos 2x$$. We have $$\sin^2 x = \frac{1-y}{2}$$, $$\cos^2 x =\frac{1+y}{2}$$. Then \begin{align} \left(\sin^2 x + \frac{1}{\sin^2 x}\right)^2 + \left(\cos^2 x + \frac{1}{\cos^2 x}\right)^2 &= \left(\frac{1-y}{2} + \frac{2}{1-y}\right)^2 + \left(\frac{1+y}{2} + \frac{2}{1+y}\right)^2 = \\ &= \frac{y^6+7y^4-y^2+25}{2(1-y^2)^2} = \\ &= \frac{25}{2} + \frac{y^2(y^4-18y^2+49)}{2(1-y^2)^2} = \\ &= \frac{25}{2} + y^2\Big(\frac{1}{2} + \frac{8}{1-y^2} + \frac{16}{(1-y^2)^2}\Big) \end{align} Since $$y^2 \le 1$$, the expression in the brackets is strictly positive.

Here is a solution using AM-GM and the double-angle formulae

• $$(\star):\sin^2 x = \frac{1-\cos 2x}{2}$$, $$\cos^2 x = \frac{1+\cos 2x}{2}$$

$$\begin{eqnarray*}\left ( \sin^2 x + \frac{1}{\sin^2 x} \right )^2 + \left ( \cos^2x + \frac{1}{\cos^2x} \right )^2 & \stackrel{AM-GM}{\geq} & 2\left ( \sin^2x + \frac{1}{\sin^2x} \right ) \left ( \cos^2x+ \frac{1}{\cos^2x} \right ) \\ & = & 2\frac{(\sin^4 x + 1)(\cos^4 x + 1)}{\sin^2 x \cdot \cos^2 x}\\ & \stackrel{AM-GM}{\geq} & 8\frac{(\sin^4 x + 1)(\cos^4 x + 1)}{(\sin^2 x + \cos^2 x)^2}\\ & \stackrel{\star}{=} & 8\left(\frac{(1-\cos 2x)^2}{4}+1 \right)\left(\frac{(1+\cos 2x)^2}{4}+1 \right)\\ & = & \frac{1}{2}(5+\cos^2 2x - 2\cos 2x)(5+\cos^2 2x + 2\cos 2x)\\ & = & \frac{1}{2}(25+\cos^4 2x +6 \cos^2 2x)\\ & \geq & \frac{25}{2} \end{eqnarray*}$$ Note that equality is attained for $$\cos 2x = 0 \Leftrightarrow \cos^2 x = \sin ^2 x$$, where the last condition is required to produce equality for AM-GM.

Another method: $$\left ( \sin^2(x) + \frac{1}{\sin^2(x)} \right )^2 + \left ( \cos^2(x) + \frac{1}{\cos^2(x)} \right )^2=\\ \sin ^4x+\cos ^4x+4+\frac1{\sin^4 x}+\frac1{\cos^4 x}=\\ (\sin ^2x+\cos ^2 x)^2-2\sin^2x\cos^2x+4+\frac{(\sin ^2x+\cos ^2 x)^2-2\sin^2x\cos^2x}{\sin^4x\cos^4x}=\\ 1-\frac12\sin^2 2x+4+\frac{1-\frac12\sin 2x}{\frac1{16}\sin^4 2x}=\\ \left(1-\frac12\sin^2 2x\right)\left(1+\frac{16}{\sin^4 2x}\right)+4\ge \\ \left(1-\frac12\cdot 1\right)\left(1+\frac{16}{1^2}\right)+4=12.5.$$ Note: $$1-\frac12\sin ^2 2x>0$$, so $$\sin^2 2x$$ should be maximized to $$1$$.

The following is not completely rigourous, but can be made so with little effort.

An obvious trigonometric identity is $$\sin^2 x+\cos^2 x=1$$. Using this, put $$y=\sin^2 x$$ to obtain a function of the form

$$f(y)=(y+\frac{1}{y})^2+(1-y+\frac{1}{1-y})^2$$ defined on the open interval $$(0,1)$$. This function is symmetric around $$y=\frac{1}{2}$$, because $$f(y)=f(1-y)$$. Also, $$\lim_{y\to 0^+}f(y)=\lim_{y\to 1^-}f(y)=+\infty$$. You can use a bit of calculus to convince yourself that $$f$$ decreases in $$(0,1/2)$$ and increases in $$(1/2,1)$$ to conclude that there is a minimum (unique, actually), when $$y=1/2$$.

AM-GM says $$a^2+b^2\geq 2ab$$ with equality when $$a=b.$$ Thus we cannot obtain a lower value than which gives GM.
With a bit of algebra we get that the equality really occurs, and this is iff $$|\sin x|=|\cos x|,$$ from where the minimum $$2.5^2+2.5^2=12.5$$