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Find the minimum value of \(\sin x + \csc x)^2 + (\cos x + \sec x)^2\) for \(0 < x < \frac{\pi}{2}.\)

 Aug 6, 2022
 #2
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minimum value is 9

 Aug 6, 2022
 #3
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Hello Guest!

Remember: 

\(sin^2 x +cos^2 x = 1 \\ csc (x) = \frac{1}{sin (x)} \\ sec(x) = \frac{1}{cos (x)}\)

So:

\((sin(x)+csc(x))^2+(cos(x)+sec(x))^2 \\ =sin^2(x)+\frac{1}{sin^2(x)}+2sin(x)\frac{1}{sin(x)}+cos^2(x)+\frac{1}{sin^2(x)}+2cos(x)\frac{1}{cos(x)} \\ =(sin^2(x)+cos^2(x))+2+2+(\frac{1}{sin^2(x)}+\frac{1}{cos^2(x)}) \\ =1+4+\frac{sin^2(x)+cos^2(x)}{sin^2(x)cos^2(x)} =5+\frac{1}{(sin(x)cos(x))^2}\)

Now, we want the minimum value, so we have to minimize: \(\frac{1}{(sin(x)cos(x))^2}\) or equivalently, maximize: \((sin(x)cos(x))^2\)

Recall: \(sin(2x)=2sin(x)cos(x) \implies sin(x)cos(x)=\frac{1}{2}sin(2x)\)

Hence, we want to maximize \(\frac{1}{4}sin^2(2x)\)

\(0 < sin^2(2x) \le 1\)

So, the upper bound is 1, which is satisfied when \(x=\frac{\pi}{4}\)

So, the maximum value of  \(\frac{1}{4}sin^2(2x)\) is: \(\dfrac{1}{4}*(1)=\dfrac{1}{4}\)

 

Therefore, the minimum value of \(\frac{1}{(sin(x)cos(x))^2}\) is \(\frac{1}{\frac{1}{4}}=4\)

Thus, the minimum value of the whole expression in the question is: 

\(5+4=9\) , as guest has found below.

I hope this helps!


 

 Aug 7, 2022

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