Problem 1
Find the minimum value of the following trigonometric function:
|sinx + cosx + tanx + cotx + secx + cscx|
Here, x is any real number except an integral multiple of π/2.
Here’s a simulation that shows the graph of this function.
Try dragging the point around. When does the function achieve the minimum value? What’s this minimum value? How can you prove that?
Solution
We’ll start with a common pitfall.
Using the AM-GM inequality, we’ll get:
(sinx + cosx + tanx + cotx + secx + cscx)/6 ≥ (sinx.cosx.tanx.cotx.secx.cscx)^(1/6)
The RHS simplifies to 1. And, we’ll get:
(sinx + cosx + tanx + cotx + secx + cscx)/6 ≥ 1
⇒ sinx + cosx + tanx + cotx + secx + cscx ≥ 6
The above deceiving inequality might lead you to the incorrect conclusion, i.e., the minimum value is 6.
But it isn’t 6! Why? That’s for you to think. I suggest you explore the AM-GM inequality a bit.
Now, another way to proceed here is to convert each ratio in terms of a single trigonometric ratio, say sinx.
Later, if we substitute sinx as y, the original expression becomes the function of a single variable y – much easier to handle than six differently behaving functions.
That’s all I have to say for now. Why don’t you try it from here on?
Were you able to solve it using the above hint? Or, did you solve using any other method? Or, do you need some more help? You can reach out at the Telegram group.
Problem Source: Putnam Competition
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