# How to correctly find the value of theta for which $\frac{a}{\cos\theta}+\frac{b}{\sin\theta}$ is minimum?

I was solving a question which required me to find the value of $$\theta$$ for which the expression $$\frac{a}{\cos\theta}+\frac{b}{\sin\theta}$$ has its minimum value. Given that, $$a=3\sqrt3, b=1$$.

Note: It was a fault from my end for not including the values of $$a,b$$ which were given in the question. I apologize for that.

Since this is a fairly trivial task, I proceeded by using AM-GM inequality which yields the following:

### Using AM-GM inequality

$$\frac{a}{\cos\theta}+\frac{b}{\sin\theta} \ge 2\sqrt{\frac{ab}{\cos\theta\sin\theta}}\\ \implies \frac{a}{\cos\theta}+\frac{b}{\sin\theta} \ge 2\sqrt{\frac{2ab}{\sin2\theta}}$$ Since, we're minising the given expression, therefore $$\sin2\theta$$ should have maximum value, i.e. $$\sin2\theta=1$$. Therefore, $$\theta=\dfrac{\pi}{4}$$.

### Using calculus

$$\text{Let }f(\theta)=a\sec\theta+b\csc\theta \\ \therefore f'(\theta)=a\sec\theta\tan\theta-b\csc\theta\cot\theta \\ \text{For mininma, }f(\theta)=0, \implies a\sec\theta\tan\theta=b\csc\theta\cot\theta \\ \text{or, }\tan^3\theta=\frac{b}{a}=\frac{1}{3\sqrt3} \\ \implies \theta=\frac{\pi}{6}$$ Why does solving this problem using derivatives yield $$\theta=\dfrac{\pi}{6}$$. Why is this happening? What should I look out for in the future while deciding whether to use AM-GM or calculus?

• Please show your work solving using derivatives; doesn't $\theta$ depend on $a$ and $b$? Apr 8, 2019 at 14:11
• @J.W.Tanner Done! Apr 8, 2019 at 14:28
• Thanks for clarifying; you have the correct answer using calculus now Apr 8, 2019 at 14:37
• Simply, you cannot apply AM-GM inequality here. This applies to means. You need calculus for a correct answer.
– Jon
Apr 8, 2019 at 14:54

Even if $$f(\theta)\ge g(\theta)$$ for all $$\theta$$, $$f$$ may attain its least possible value for some $$\theta$$ that doesn't minimize $$g$$. Consider for instance the inequality (which holds on the whole real line) $$\theta^2\ge \frac12\theta^2-\theta-1$$ The LHS is minimized for $$\theta=0$$, the RHS for $$\theta=1$$.

So it's not really a matter of choice of approach. It's a matter of basic logic.

• Thinking about this problem also led me to this conclusion, but I still want to know when to rely upon either of the methods. That's also what my question aims at! Apr 8, 2019 at 14:31
• Do both and the one that works faster is the one you should have used first.
– user562983
Apr 8, 2019 at 14:32
• This when you're young. When you're old, use the one that has worked faster most of the times and, if it doesn't work, cry like an old man.
– user562983
Apr 8, 2019 at 14:34
• Sadly, I don't have that freedom. I'm preparing for an exam, hence time is limited. I'd like to save as much time as I can, hence even the thought of doing multiple methods for a single problem is out of the question. Apr 8, 2019 at 14:34

Let $$y=a\sec t+b\csc t$$

$$\dfrac{dy}{dt}=a\sec t\tan t-b\csc t\cot t=\dfrac{a\sin^3t-b\cos^3t}{\cos^2t\sin^2t}$$

For extreme values of $$y,f'(t)=0$$

$$\implies\tan^3t=\dfrac ba\iff\dfrac a{\cos^3t}=\dfrac b{\sin^3t}=\pm\sqrt{a^{2/3}+b^{2/3}}$$

How have you found $$t=\dfrac\pi8?$$

• The values for a,b were provided in the question. I have now included them in the question. Sorry for not mentioning them beforehand. Apr 8, 2019 at 14:29
• @Utkarsh, So, put those values in $f'(t)$ Apr 8, 2019 at 14:38
• The answer comes $\frac{\pi}{6}$ which is different from the one given by AM-GM inequality. Apr 8, 2019 at 14:40