Minimum value Of trigonometry expression

Question

FIND THE MIN VALUE OF 4 cosec^2 x + 9 sin^2 x ? Please explain by both calculus and non-calculus methods ?

Answer 1

Another way using AM-GM inequality: for positive $a,b$

$$ \frac{a+b}{2} \geq \sqrt{ab} $$

Notice

$$ 4 \csc^2 x + 9 \sin^2 x = (2 \csc x)^2 + (3 \sin x)^2 \geq 2 \cdot 2 \csc x \cdot 3 \sin x = 12 \frac{1}{ \sin x}{\sin x} = 12$$

Therefore, we conclude that the minimum value is $12$

Answer 2

$$(2\csc x)^2+(3\sin x)^2=(2\csc x-3\sin x)^2+2\cdot 2\csc x\cdot3\sin x\ge2\cdot 2\cdot3$$

Answer 3

Hint :

consider taking derivative of $4 \csc^2 x + 9 \sin^2 x $

Answer 4

Particularly if you are in a mood to differentiate, it is easier to solve the equivalent problem of minimizing $\dfrac{4}{t}+9t$, where $0\lt t\le 1$.