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Published byFreddie Mattox Modified over 2 years ago

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GRAPHS OF Secant and Cosecant Functions

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For the graph of y = f(x) = sec x we'll take the reciprocal of the cosine values. x cos x y = sec x x y 1 - 1

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y = f(x) = sec x Choose more values. x cos x y = sec x Since the secant is the reciprocal of the cosine, the cosine graph will help graph the secant graph. x y 1 - 1

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The vertical lines are not part of the graph but are where the secant is undefined (which is where the cosine was 0) Let's look over a few periods at the graph of y =sec x Let's add in the graph of the cosine function so you can see how if you graph it, you can then easily use it to graph the secant.

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For the graph of y = f(x) = csc x we'll take the reciprocals of the sine values. x sin x y = csc x When we graph these rather than plot points after we see this, we'll use the sine graph as a sketching aid and then get the cosecant graph. x y 1 - 1

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y = f(x) = csc x choose more values x sin x y = csc x We'll use the sine graph as the sketching aid. x y 1 - 1 When the sine is 0 the cosecant will have an asymptote.

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Again the vertical lines are not part of the graph but are where the cosecant is undefined (which is where the sine was 0 so taking the reciprocal, these values are not in the domain of the cosecant function. Let's add in the graph of the sine function so you can see how if you graph it, you can then easily use it to graph the cosecant. Let's look over a few periods at the graph of y = csc x

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We’ll use the reciprocal function as our sketching aid so first we’ll graph Amplitude? Period? Phase Shift? 1 2 /1 2 /4 Start with basic cosine graphChange amplitude to 2No change in period Everything to the right /4 Now use this to graph secant, the reciprocal function. Everywhere there is an x intercept for cosine there will be an asymptote and everywhere there is a max or min will be the turning point for the secant function. Let’s remove all of the sketching aids now and have a look at the secant graph.

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