Question

range of f(x) = [ sin x + [ cos x + [ tan x + [sec x]]]] if x belongs to (0,pi/4)

Answer 1

{1}

Answer 2

To find the range of $f(x) = [ \sin x + [ \cos x + [ \tan x + [\sec x]]]]$ for $x \in (0, \frac{\pi}{4})$, we need to analyze the values of each trigonometric function within the given interval and then apply the greatest integer function (floor function) from the innermost bracket outwards.

Step 1: Determine the range of each trigonometric function for $x \in (0, \frac{\pi}{4})$

• $\sin x$: As $x$ increases from $0$ to $\frac{\pi}{4}$, $\sin x$ increases from $\sin(0)=0$ to $\sin(\frac{\pi}{4})=\frac{1}{\sqrt{2}}$. So, $0 < \sin x < \frac{1}{\sqrt{2}} \approx 0.707$.
• $\cos x$: As $x$ increases from $0$ to $\frac{\pi}{4}$, $\cos x$ decreases from $\cos(0)=1$ to $\cos(\frac{\pi}{4})=\frac{1}{\sqrt{2}}$. So, $\frac{1}{\sqrt{2}} < \cos x < 1$, i.e., $0.707 < \cos x < 1$.
• $\tan x$: As $x$ increases from $0$ to $\frac{\pi}{4}$, $\tan x$ increases from $\tan(0)=0$ to $\tan(\frac{\pi}{4})=1$. So, $0 < \tan x < 1$.
• $\sec x$: Since $\sec x = \frac{1}{\cos x}$, and $\frac{1}{\sqrt{2}} < \cos x < 1$, taking reciprocals reverses the inequality: $\frac{1}{1} < \frac{1}{\cos x} < \frac{1}{1/\sqrt{2}}$. So, $1 < \sec x < \sqrt{2} \approx 1.414$.

Step 2: Evaluate the greatest integer function for each term

• $[ \sin x ]$: Since $0 < \sin x < 0.707$, $[\sin x] = 0$.
• $[ \cos x ]$: Since $0.707 < \cos x < 1$, $[\cos x] = 0$.
• $[ \tan x ]$: Since $0 < \tan x < 1$, $[\tan x] = 0$.
• $[ \sec x ]$: Since $1 < \sec x < 1.414$, $[\sec x] = 1$.

Step 3: Evaluate the nested greatest integer functions from inside out

The function is $f(x) = [ \sin x + [ \cos x + [ \tan x + [\sec x]]]]$

1. Innermost bracket: $[\sec x]$ From Step 2, $[\sec x] = 1$. So, $f(x) = [ \sin x + [ \cos x + [ \tan x + 1]]]$

2. Next bracket: $[ \tan x + 1 ]$ We know $0 < \tan x < 1$. Adding 1 to all parts of the inequality: $0+1 < \tan x + 1 < 1+1 \implies 1 < \tan x + 1 < 2$. Therefore, $[ \tan x + 1 ] = 1$. (Alternatively, using the property $[y+n] = [y]+n$ for integer $n$: $[\tan x + 1] = [\tan x] + 1 = 0 + 1 = 1$). So, $f(x) = [ \sin x + [ \cos x + 1]]$

3. Next bracket: $[ \cos x + 1 ]$ We know $\frac{1}{\sqrt{2}} < \cos x < 1$. Adding 1 to all parts of the inequality: $1 + \frac{1}{\sqrt{2}} < \cos x + 1 < 1 + 1 \implies 1.707 < \cos x + 1 < 2$. Therefore, $[ \cos x + 1 ] = 1$. (Alternatively, $[\cos x + 1] = [\cos x] + 1 = 0 + 1 = 1$). So, $f(x) = [ \sin x + 1]$

4. Outermost bracket: $[ \sin x + 1 ]$ We know $0 < \sin x < \frac{1}{\sqrt{2}}$. Adding 1 to all parts of the inequality: $0+1 < \sin x + 1 < \frac{1}{\sqrt{2}} + 1 \implies 1 < \sin x + 1 < 1.707$. Therefore, $[ \sin x + 1 ] = 1$. (Alternatively, $[\sin x + 1] = [\sin x] + 1 = 0 + 1 = 1$).

Thus, for all $x \in (0, \frac{\pi}{4})$, $f(x) = 1$.

The range of $f(x)$ is the set of all possible values of $f(x)$. Since $f(x)$ always evaluates to 1 in the given domain, the range is $\{1\}$.