To find minimum value of 2 csc ⁡ ( 2 x ) + sec &#x20

raulgallerjv 2022-05-23 Answered

To find minimum value of

2 \csc (2 x) + \sec (x) + \csc (x)

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tatoas9f

Answered 2022-05-24 Author has 11 answers

Method #1:
If we set

\sin x + \cos x = u, u \leq \sqrt{2}, u^{2} = 1 + 2 \sin x \cos x

we need to minimize

2 \cdot \frac{u + 1}{u^{2} - 1} = \frac{2}{u - 1}

i.e., to maximize u−1
Method #2:
As

(\sin x + \cos x)^{2} - 1^{2} = 2 \sin x \cos x

and

\sin x + \cos x = \sqrt{2} \sin (x + \frac{π}{4})

\frac{\sin x + \cos x + 1}{\sin x \cos x} = \frac{2}{\sin x + \cos x - 1} = \frac{2}{\sqrt{2} \sin (x + \frac{π}{4}) - 1}

Now for

x \in (0, \frac{π}{2}),

,

\frac{1}{\sqrt{2}} < \sin (x + \frac{π}{4}) \leq 1

Method #3:

2 \sin x \cos x = - \cos 2 (\frac{π}{4} + x) = {(\sqrt{2} \sin (x + \frac{π}{4}))}^{2} - 1^{2}

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