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To find minimum value of $$2\csc(2x)+\sec(x)+\csc(x)$$ for $x \in (0,\frac{\pi}{2})$

i converted to sin and cos so that i may reduce it to form $a\cos(x)+b\sin(x)$. I reduced it to $$\frac{\sin(x)+\cos(x)+1}{\sin(x)\cos(x)}$$

How do i proceed from here?

Thanks a ton!

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    Did you cover derivatives yet? – 2017-02-08
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    @Math_QED Nope. – 2017-02-08

2 Answers 2

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Method $\#1:$

If we set $\sin x+\cos x=u, u\le\sqrt2, u^2=1+2\sin x\cos x$

we need to minimize $$2\cdot\dfrac{u+1}{u^2-1}=\dfrac2{u-1}$$

i.e., to maximize $u-1$

Method $\#2:$

As $(\sin x+\cos x)^2-1^2=2\sin x\cos x$

and $\sin x+\cos x=\sqrt2\sin\left(x+\dfrac\pi4\right)$

$$\dfrac{\sin x+\cos x+1}{\sin x\cos x}=\dfrac2{\sin x+\cos x-1}=\dfrac2{\sqrt2\sin\left(x+\dfrac\pi4\right)-1}$$

Now for $x\in\left(0,\dfrac\pi2\right),$

$$\dfrac1{\sqrt2}<\sin\left(x+\dfrac\pi4\right)\le1$$

Method $\#3:$

$$2\sin x\cos x=-\cos2\left(\dfrac\pi4+x\right)=\left(\sqrt2\sin\left(x+\dfrac\pi4\right)\right)^2-1^2$$

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    How did you know that we have to take sinx+cosx=u to solve this? thanks – 2017-02-08
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    @Gathdi, As I know $(\sin x+\cos x)^2=\cdots?$ – 2017-02-08
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    Is there other way by using trig identities to reduce it – 2017-02-08
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    i still don't get your hint.the value is $1+sin(2x)$ – 2017-02-08
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    @Gathdi, What is $\sin2x?$ – 2017-02-08
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    sin2x=$2sinxcosx$ – 2017-02-08
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    @Gathdi, So if $\sin x+\cos x=u,\sin x\cos x=?$ – 2017-02-08
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    okay i got this.btw is this can be done by using identities and no substitution? – 2017-02-08
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/53234/discussion-between-lab-bhattacharjee-and-gathdi). – 2017-02-08
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    Is it not that maximum value of $u-1 is 2(\sqrt{2}-1)$. But answer in textbook is $2(\sqrt{2}+1)$ – 2017-02-08
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    Kindly let me know answer to my query – 2017-02-08
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    @Gathdi, Maximum value of $u-1$ is $\sqrt2-1$ So, the required minimum is $$\dfrac2{\sqrt2-1}=?$$ – 2017-02-08
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    why don't we need to calculate range for t like x in (0 , pi\2) – 2017-02-08
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    @Gathdi, $u$ attains maximum at ? – 2017-02-08
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    u attains maximum at $\sqrt{2}$ – 2017-02-08
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Using $\sin2A=2\sin A\cos A,\cos2A=\cos^2A-\sin^2A$

$$\dfrac{\sin x+\cos x+1}{\sin x\cos x}=\dfrac{2\cos\dfrac x2\left(\cos\dfrac x2+\sin\dfrac x2\right)}{2\cos\dfrac x2\sin\dfrac x2\left(\cos^2\dfrac x2-\sin^2\dfrac x2\right)}=\dfrac1{\sin\dfrac x2\left(\cos\dfrac x2-\sin\dfrac x2\right)}$$

Now $\cos\dfrac x2-\sin\dfrac x2=\sqrt2\sin\left(\dfrac\pi4-\dfrac x2\right)$

$\implies\sin\dfrac x2\left(\cos\dfrac x2-\sin\dfrac x2\right)=\sqrt2\sin\dfrac x2\sin\left(\dfrac\pi4-\dfrac x2\right)=\dfrac{\cos\left(x-\dfrac\pi8\right)-\cos\dfrac\pi8}{\sqrt2}$ which needs to be maximized with $x \in \left(0,\dfrac\pi2\right)$

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    How to maximise it ? – 2017-02-08
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    @Gathdi, When cosine attains maximum ? – 2017-02-08
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    @Gathdi, So, here we need $x-\dfrac\pi8=0$ – 2017-02-08
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    So i have to put x=pi\8 in original expression – 2017-02-08
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    When i put value of cos(pi\8) which is cos(45/2) in expression in your last step answer comes to be different – 2017-02-09
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    @Gathdi, Please write down how & what result you have reached? – 2017-02-09