Minimum value of this trigonometric function

Question

Find the minimum value of this function for $0\lt \theta \lt \pi$ $$y = 3\sin{\theta} + \text{cosec}^3\theta$$

What I want to specifically ask is about use of inequality rules in this function.

I used the rule: $$\text{if } a\lt b \text{ and } c\lt d \implies a + c\lt b + d$$

$$0\lt3\sin\theta\lt 3\text{ }\forall\text{ } 0\lt \theta \lt \pi$$ Also $$1\lt\text{cosec}^3\theta\lt \infty$$ So shouldn't these two inequations combine to give $$1\lt3\sin{\theta} + \text{cosec}^3\theta\lt\infty$$ But the correct answer is 4. Did I do any mistake or does this rule not apply here?

I'm not looking for solution to this problem, I just want to know why this method fails here.

Answer 1

Basically, you got that the sum of the function should lie in that interval. However, this tells nothing about what the minimum value of the sum can be.

An analogy: Say you are in your home then I can derive from that premise that you are in some room of your home but I can't really say which room you are in. The only thing I know is that you must be fixed in space of your house.

Similarly, this inequality doesn't really tell us how big or small the function can actually be. Only where the values of it can lie in.

For example:

$$ -2 < \sin \theta < 2$$

Is a true inequality for all $\theta$, however this does not mean that $2$ is maximum of $ \sin \theta$

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Clarifications:

Inequalities are indeed a difficult subject.. however if you boil it down to fundamentals then you can make it simple.

For example consider the set $(1,3)$ this is equivalent to the inequality:

$$ 1 < x < 3$$

To show that this problem arises even for non trignometric functions , consider the function $f(x) = -x^2$

It is pretty simple to see that:

$$ f(x) \leq 0 $$

For all $ x \in R$

However, this doesn't mean the inequality business is completely useless. There are still some nice conclusions which you can make.

For example, consider the function $ f(x) = x \ln x $, from our days in calculus we know the exact methodologies for finding the maxima and minimum.

$$ f'(x) = \ln x + 1$$

Or for the maxima or a minima,

$$ \ln x = -1$$

Which suggests $ x = \frac{1}{e}$

And, from the second derivative test, $ f''(x)$ is positive at $x=e$. Hence this means that it is a minimum occurs here, since this is the only turning point in the graph, it must be absolute minima.

Hence we can boldly make the statement that:

$$ \frac{1}{e} \leq f(x)$$

Setting a lower bound for the function.

Answer 2

For $0 \lt \theta \lt \pi,$ $cosec \theta \ge 1$ (min occurs when $\sin \theta$ is maximum i.e $\sin \theta = 1$).

Say, $cosec \theta = 1+x, $ where $x \ge 0$

$y = \frac{3}{1+x} + (1+x)^3= 1+x^3+3x(1+x)+\frac{3}{1+x} = 4 + x^3+3x(1+x)-\frac{3x}{1+x}$

$\implies y = 4 + x^3+3x(1+x)-\frac{3x}{1+x} = 4 + x^3 + \frac{3x(x^2+2x)}{1+x} \ge 4$

The min value of $4$ occurs when $x = 0$ (i.e $cosec\theta = 1 = \sin\theta$).

Answer 3

The property $y>1$ is true, but it does not give a minimal value because you need also to show that the minimal value occurs, as in the following reasoning.

By AM-GM $$y\geq4\sqrt[4]{\sin^3x\cdot\frac{1}{\sin^3x}}=4.$$ The equality occurs for $x=\frac{\pi}{2},$ which says that we got a minimal value.