Math  /  Trigonometry

Question18. Using sketches of f(x)=secxf(x)=\sec x and g(x)=cscxg(x)=\csc x, determine where secx>cscx\sec x>\csc x for 0rx2πr0^{r} \leq x \leq 2 \pi^{r}. State your answer using interval notation. Include any asymptotes in your sketches.

Studdy Solution

STEP 1

1. We are analyzing the functions f(x)=secx f(x) = \sec x and g(x)=cscx g(x) = \csc x .
2. The domain of interest is 0x2π 0 \leq x \leq 2\pi .
3. We need to determine where secx>cscx \sec x > \csc x .

STEP 2

1. Understand the behavior and properties of secx \sec x and cscx \csc x .
2. Identify the asymptotes and critical points of both functions.
3. Sketch the graphs of secx \sec x and cscx \csc x .
4. Determine the intervals where secx>cscx \sec x > \csc x .
5. State the solution in interval notation.

STEP 3

The function secx=1cosx \sec x = \frac{1}{\cos x} and is undefined where cosx=0 \cos x = 0 , which occurs at x=π2+nπ x = \frac{\pi}{2} + n\pi for integers n n .
The function cscx=1sinx \csc x = \frac{1}{\sin x} and is undefined where sinx=0 \sin x = 0 , which occurs at x=nπ x = n\pi for integers n n .

STEP 4

Identify the asymptotes: - secx \sec x has vertical asymptotes at x=π2 x = \frac{\pi}{2} and x=3π2 x = \frac{3\pi}{2} . - cscx \csc x has vertical asymptotes at x=0,π,2π x = 0, \pi, 2\pi .

STEP 5

Sketch the graphs: - secx \sec x oscillates between its asymptotes, with branches opening upwards from x=0 x = 0 to x=π2 x = \frac{\pi}{2} and from x=3π2 x = \frac{3\pi}{2} to x=2π x = 2\pi , and downwards from x=π2 x = \frac{\pi}{2} to x=3π2 x = \frac{3\pi}{2} . - cscx \csc x oscillates between its asymptotes, with branches opening upwards from x=0 x = 0 to x=π x = \pi and from x=π x = \pi to x=2π x = 2\pi .

STEP 6

Determine where secx>cscx \sec x > \csc x : - Compare the values of secx \sec x and cscx \csc x between the asymptotes. - Between x=0 x = 0 and x=π2 x = \frac{\pi}{2} , secx>cscx \sec x > \csc x . - Between x=π2 x = \frac{\pi}{2} and x=π x = \pi , cscx>secx \csc x > \sec x . - Between x=π x = \pi and x=3π2 x = \frac{3\pi}{2} , secx>cscx \sec x > \csc x . - Between x=3π2 x = \frac{3\pi}{2} and x=2π x = 2\pi , cscx>secx \csc x > \sec x .

STEP 7

State the solution in interval notation: The intervals where secx>cscx \sec x > \csc x are (0,π2)(π,3π2) \left(0, \frac{\pi}{2}\right) \cup \left(\pi, \frac{3\pi}{2}\right) .
The solution is:
(0,π2)(π,3π2) \boxed{\left(0, \frac{\pi}{2}\right) \cup \left(\pi, \frac{3\pi}{2}\right)}

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