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# help

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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

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minimum value is 9

Aug 6, 2022
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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