Trigonometry: find minum of y=Tan(x)^p+Cot(x)^q

[hadi amiri 4]

Homework Statement

suppose p and q are positive rational numbers with the condition : 0<x<Pi/2
find the minimum y=Tan(x)^p+Cot(x)^q
Note:with trignonometry

Homework Equations

The Attempt at a Solution

 

[HallsofIvy]

When you posted this before, you were given a pretty good explanation of how to solve this problem. Have you tried that yourself?

 

[Architeuthis Dux]

To find the minimum value of y, we can use the properties of trigonometric functions and the concept of optimization. First, we can rewrite the equation as y = sin(x)^p/cos(x)^p + cos(x)^q/sin(x)^q. Then, we can use the fact that sin(x) and cos(x) are bounded by -1 and 1 to rewrite the equation as y = (sin(x)^p + cos(x)^q)/(cos(x)^p * sin(x)^q).

Next, we can use the AM-GM inequality, which states that the arithmetic mean of a set of numbers is greater than or equal to the geometric mean of the same set of numbers. Applying this to our equation, we get y >= 2^(p+q)/cos(x)^p * sin(x)^q.

Since 0<x<Pi/2, cos(x) and sin(x) are both positive, and the minimum value of the expression will occur when cos(x)^p * sin(x)^q is minimized. This happens when cos(x)^p = sin(x)^q, which can be rewritten as cos(x)/sin(x) = (1/cos(x))^(p/q).

Using the fundamental trigonometric identity, cos(x)/sin(x) = cot(x), we get cot(x) = (1/cos(x))^(p/q) or cot(x) = (sin(x)/cos(x))^(p/q).

Since p and q are positive rational numbers, we can rewrite them as fractions, p = a/b and q = c/d, where a, b, c, and d are positive integers. Therefore, the equation becomes cot(x) = (sin(x)/cos(x))^(a/bc/d).

Using the power property of trigonometric functions, we can rewrite this as cot(x) = (sin(x)^(a/b))/(cos(x)^(c/d)).

Now, we can use the AM-GM inequality again to find the minimum value of cot(x). The geometric mean of sin(x)^(a/b) and cos(x)^(c/d) is equal to (sin(x)^(a/b) * cos(x)^(c/d))^(1/2).

Using the fact that sin(x)^2 + cos(x)^2 = 1, we get that the minimum value of cot(x) is 2^(-1/2), which