# Is there a better way to find the minimum value of $\frac{1}{\sin \theta \cos \theta \left ( \sin \theta + \cos \theta \right )}$?

To find the minimum value of $$\frac{1}{\sin \theta \cos \theta \left ( \sin \theta + \cos \theta \right )}$$

I can see that I can convert it to

$$\frac{\sqrt{2}}{\sin (2 \theta) \sin \left ( \theta + \frac{\pi}{4} \right )}$$

And from here I can convert it to a pair of cosecant functions and then proceed with the usual product rule, solving that out with zero etc. Once that is all done, we can find that the function attains a minimum at $\theta= \frac{\pi}{4}$ and this minimum value is $\sqrt{2}$.

However, I was wondering if there was perhaps a better way of finding the minimum value of this function that does NOT require heavy amounts of algebra bashing. Perhaps by some means of observation.

So for example, one could find the maximum of $\sin \theta$ by realising that it oscillates between $1$ and $-1$. Could a similar idea be used to find the minimum of this function?

• What if $\theta \to 0$ from the left? Wouldn't this tend to $-\infty$ and have no minimum? – jameselmore Sep 29 '15 at 16:24
• if f(x) has an extremum at x = x0, then 1/f(x) will also have an extremum there (f(x) = 0 being an exceptional case). Also, you can consider log[f(x)]. – Count Iblis Sep 29 '15 at 16:24
• @CountIblis Your comment is sufficient for an answer. – Trogdor Sep 29 '15 at 16:26

If $a^2+b^2=1$, then we can prove that $ab(a+b) \leq \frac{1}{\sqrt{2}}$ using: $ab \leq \frac{1}{2}(a^2+b^2)=\frac{1}{2}$ and $(a+b)^2=1+2ab \leq 2$. For equality to be attained, it is necessary that $a=b$.

• Just so beautiful. – Trogdor Sep 29 '15 at 16:33

Another approach:

Let $$\sin x+\cos x=p$$ then $p \in [-\sqrt{2} \: \sqrt{2}]$

Also $$\sin x \:\cos x=\frac{p^2-1}{2}$$ So we need to find Max and Min values of the function

$$g(p)=\frac{2}{p^3-p}=\frac{2}{f(p)}$$ in the interval $p \in [-\sqrt{2}\:\sqrt{2}]$

it will be evident from the Graph of $f(p)$ that in the above

interval $f(p)$ is Min at $p=-\sqrt{2}$ and is Max at $p=\sqrt{2}$

So

$\max(f(p))=\sqrt{2}$ and $\min(f(p))=-\sqrt{2}$

Thus

$\displaystyle \max(g(p))=\frac{2}{\sqrt{2}}=\sqrt{2}$ and Similarly $\displaystyle \min(g(p))=-\sqrt{2}$

Hint: Use $$\frac{1}{\sin \theta \cos \theta \left ( \sin \theta + \cos \theta \right )} = \frac {4 \cos \theta - 4 \sin \theta} {\sin {4 \theta}}.$$

Sine and Cosine are bounded. The minimum value is the one for which sine and Cosine have both the maximum value namely for $\theta = \frac{\pi}{4}$, so you will have $\sin\theta = \cos\theta = \frac{\sqrt{2}}{2}$

Substituting and you'll get

$$f(\theta) = \frac{1}{\left(\frac{2}{4}\left(2\frac{\sqrt{2}}{2}\right)\right)} = \frac{2}{\sqrt{2}} = \sqrt{2}$$

Thus the minimum is $- \sqrt{2}$ and the max is $\sqrt{2}$