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Find the range of the function $$f(x)=\sqrt{\log(\cos(\sin(x))}.$$

What I did was:

$$\log(\cos(\sin(x))) \geq 0$$ Then $$\log_e\cos(\sin(x)) \geq 0$$ The next step of the problem is $$\cos(\sin(x)) \geq e^0$$ This is the part I didn’t understand. Please explain it to me.

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Note that $\sqrt{\log(\cos(\sin(x)))}$ is defined if and only if $\cos(\sin(x))\geq 1$ otherwise we have the square root of a negative number. Now since the cosine function is upper bounded by $1$, $\cos(\sin(x))$ has to be $1$ (such equality holds for example when $x=0$). So what is the range of the given function?

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  • $\begingroup$ Why can it be greater than or equal to zero? Why does it have to be greater than or equal to one? That’s the part I didn’t understand $\endgroup$ – Aditya Jul 20 at 15:18
  • $\begingroup$ If $0<t<1$ then $\log(t)<0$ and $\sqrt{\log(t)}$ is not defined. Hence $t=\cos(\sin(x))\geq 1$. $\endgroup$ – Robert Z Jul 20 at 15:20
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Okay. $e > 1$.

So if $x > 0$ then $e^x > 1$ and $e^0 = 1$ and if $y < 0$ then $e^{y} = e^{-|y|} = \frac 1{e^{|y|} }< 1$ and so $0 < e^y < 1$.

So $e^x > 0$ always. $0 < e^x < 1$ if and only if $x < 0$. and $e^x = 1$ if and only if $x = 0$ and $e^x > 1$ if and only if $x > 0$.

Claim: $a < b$ if and only if $e^a < e^b$.

Pf: If $a < b$ then $b-a > 0$ so $e^{b-a} > 1$ and so

$e^b = e^{a + (b-a)} = e^a*e^{b-a} > e^a* 1 = e^a$.

If $e^a < e^b$ then $1 < \frac {e^b}{e^a} = e^{b-a}$. So $b-a > 0$ and $b > a$.

Claim 2: If $a > 0$ and $b > 0$, then $a < b \iff \ln a < \ln b$.

Pf:

If $a > 0$ then $a = e^{\ln a}$ and if $b > 0$ then $b = e^{\ln b}$.

So $a < b$ if and only if $e^{\ln a} < e^{\ln b}$ if and only if $\ln a < \ln b$

.......

And now we can do your problem.

$\sqrt{\ln(\cos(\sin x))} $ only exists if

$\ln(\cos(\sin x)) \ge 0$. This occurs if and only if

$\cos(\sin x) \ge e^0 = 1$.

But $-1 \le \cos w \le 1$ so the only way this can happen is if $\cos(\sin x) = 1$.

Now $\cos w = 1$ can only happen if $w = 0$.

So $\sin x = 0$. That occurs if $x = 0$ or $x = 180^{\circ}$.

So ....

$x = 0$ or $180^{\circ}$

$\sin x = 0$

$\cos(\sin x) = 1$

$\log (\cos(\sin x)) = 0$

$\sqrt{\log (\cos(\sin x))} = 0$

and the range is $\{0\}$.

The end.

BTW was $\log$ suppose to be $\log_{10}$ or $\log_e = \ln$? It doesn't actually matter. THe answer will be the same.

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  • $\begingroup$ I didn’t figure out cos(sinx) greater than or equal to 1. I didn’t understand that part which involves the base e and e^0. I want that part explained $\endgroup$ – Aditya Jul 20 at 15:19
  • $\begingroup$ Ohhh... okay... $\endgroup$ – fleablood Jul 20 at 15:22

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