Unformatted text preview: is case is 2x âˆ’ Ï€ . Loosely
stated, the argument of a trigonometric function is the expression â€˜insideâ€™ the function.
Example 10.5.1. Graph one cycle of the following functions. State the period of each.
1. f (x) = 3 cos Ï€ xâˆ’Ï€
2 +1 2. g (x) = 1
2 sin(Ï€ âˆ’ 2x) + 3
2 Solution.
âˆ’
1. We set the argument of the cosine, Ï€x2 Ï€ , equal to each of the values: 0,
solve for x. We summarize the results below. a
0
Ï€
2 Ï€
3Ï€
2 2Ï€ Ï€xâˆ’Ï€
=a
2
Ï€xâˆ’Ï€
=0
2
Ï€xâˆ’Ï€
=Ï€
2
2
Ï€xâˆ’Ï€
=Ï€
2
Ï€xâˆ’Ï€
Ï€
= 32
2
Ï€xâˆ’Ï€
= 2Ï€
2 Ï€
2, Ï€, 3Ï€
2, 2Ï€ and x
1
2
3
4
5 âˆ’
Next, we substitute each of these x values into f (x) = 3 cos Ï€x2 Ï€ + 1 to determine the
corresponding y values and connect the dots in a pleasing wavelike fashion.
y x f (x) (x, f (x)) 4
3 1 4 (1, 4) 2 2 1 (2, 1) 1 3 âˆ’2 (3, âˆ’2) 4 1 (4, 1) 5 4 (5, 4) 1 2 3 4 5 x âˆ’1
âˆ’2 One cycle of y = f (x). One cycle is graphed on [1, 5] so the period is the length of that interval which is 4.
2. Proceeding as above, we set the argument of the sine, Ï€ âˆ...
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 Fall '13
 Wong
 Algebra, Trigonometry, Cartesian Coordinate System, The Land, The Waves, RenÃ© Descartes, Euclidean geometry

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