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

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.

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?

• 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 – Aditya Jul 20 '19 at 15:18
• If $0<t<1$ then $\log(t)<0$ and $\sqrt{\log(t)}$ is not defined. Hence $t=\cos(\sin(x))\geq 1$. – Robert Z Jul 20 '19 at 15:20

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.

• 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 – Aditya Jul 20 '19 at 15:19
• Ohhh... okay... – fleablood Jul 20 '19 at 15:22