Graph y=csc(x/2)

Problem

y=csc(x/2)

Solution

1. Identify the parent function and its properties. The function is y=csc(θ) which is the reciprocal of sin(θ) It has vertical asymptotes where sin(θ)=0

2. Determine the period of the function. The period of csc(b*x) is calculated as (2*π)/|b| Here, b=1/2

Period=(2*π)/(1/2)=4*π

3. Locate the vertical asymptotes by setting the argument of the cosecant function equal to multiples of π

x/2=n*π

x=2*n*π

For one full cycle starting at x=0 the asymptotes occur at x=0 x=2*π and x=4*π

4. Find key points for the local minimums and maximums. These occur halfway between the asymptotes.
   For the first interval (0,2*π) the midpoint is x=π

y=csc(π/2)=1

For the second interval (2*π,4*π) the midpoint is x=3*π

y=csc((3*π)/2)=−1

5. Sketch the graph by drawing the vertical asymptotes at x=…,−2*π,0,2*π,4*π,… and plotting the relative extrema. Draw U-shaped curves opening upward from (2*π*n+π,1) and downward from (2*π*n+3*π,−1)

Final Answer

y=csc(x/2)

The graph has a period of 4*π vertical asymptotes at x=2*n*π local minima at (π+4*n*π,1) and local maxima at (3*π+4*n*π,−1)

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