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8.1 Graphs of the Sine and Cosine Functions
8.2 Graphs of the Other Trigonometric Functions
8.1 Section Exercises
The sine and cosine functions have the property that f(x+P)=f(x) for a certain P. This means that the function values repeat for every P units on the x-axis.
The absolute value of the constant A (amplitude) increases the total range and the constant D (vertical shift) shifts the graph vertically.
At the point where the terminal side of t intersects the unit circle, you can determine that the sint equals the y-coordinate of the point.
amplitude: 23; period: 2π; midline: y=0; maximum: y=23 occurs at x=0; minimum: y=−23 occurs at x=π; for one period, the graph starts at 0 and ends at 2π
amplitude: 4; period: 2π; midline: y=0; maximum y=4 occurs at x=π2; minimum: y=−4 occurs at x=3π2; one full period occurs from x=0 to x=2π
amplitude: 1; period: π; midline: y=0; maximum: y=1 occurs at x=π; minimum: y=−1 occurs at x=π2; one full period is graphed from x=0 to x=π
amplitude: 3; period: π4; midline: y=5; maximum: y=8 occurs at x=0.12; minimum: y=2 occurs at x=0.516; horizontal shift: −4; vertical translation 5; one period occurs from x=0 to x=π4
amplitude: 5; period: 2π5; midline: y=−2; maximum: y=3 occurs at x=0.08; minimum: y=−7 occurs at x=0.71; phase shift: −4; vertical translation: −2; one full period can be graphed on x=0 to x=2π5
amplitude: 1 ; period: 2π; midline: y=1; maximum: y=2 occurs at x=2.09; maximum: y=2 occurs at t=2.09; minimum: y=0 occurs at t=5.24; phase shift: −π3; vertical translation: 1; one full period is from t=0 to t=2π
amplitude: 1; period: 4π; midline: y=0; maximum: y=1 occurs at t=11.52; minimum: y=−1 occurs at t=5.24; phase shift: −10π3; vertical shift: 0
A linear function is added to a periodic sine function. The graph does not have an amplitude because as the linear function increases without bound the combined function h(x)=x+sinx will increase without bound as well. The graph is bounded between the graphs of y=x+1 and y=x-1 because sine oscillates between −1 and 1.
The graph is symmetric with respect to the y-axis and there is no amplitude because the function’s bounds decrease as |x| grows. There appears to be a horizontal asymptote at y=0 .
8.2 Section Exercises
Since y=cscx is the reciprocal function of y=sinx, you can plot the reciprocal of the coordinates on the graph of y=sinx to obtain the y-coordinates of y=cscx. The x-intercepts of the graph y=sinx are the vertical asymptotes for the graph of y=cscx.
- ⓐ (−π2,π2);
- ⓑ
- ⓒ x=−π2 and x=π2; the distance grows without bound as |x| approaches π2 —i.e., at right angles to the line representing due north, the boat would be so far away, the fisherman could not see it;
- ⓓ3; when x=−π3, the boat is 3 km away;
- ⓔ 1.73; when x=π6, the boat is about 1.73 km away;
- ⓕ 1.5 km; when x=0
- ⓐ h(x)=2tan(π120x);
- ⓑ
- ⓒ h(0)=0: after 0 seconds, the rocket is 0 mi above the ground; h(30)=2: after 30 seconds, the rockets is 2 mi high;
- ⓓAs x approaches 60 seconds, the values of h(x) grow increasingly large. The distance to the rocket is growing so large that the camera can no longer track it.
8.3 Section Exercises
The function y=sinx is one-to-one on [−π2,π2]; thus, this interval is the range of the inverse function of y=sinx, f(x)=sin−1x. The function y=cosx is one-to-one on [0,π]; thus, this interval is the range of the inverse function of y=cosx,f(x)=cos−1x.
In order for any function to have an inverse, the function must be one-to-one and must pass the horizontal line test. The regular sine function is not one-to-one unless its domain is restricted in some way. Mathematicians have agreed to restrict the sine function to the interval [−π2,π2] so that it is one-to-one and possesses an inverse.
True . The angle, θ1 that equals arccos(−x), x>0, will be a second quadrant angle with reference angle, θ2, where θ2 equals arccosx, x>0. Since θ2 is the reference angle for θ1, θ2=π−θ1 and arccos(−x) = π−arccosx-
Review Exercises
The graphs are not symmetrical with respect to the line y=x. They are symmetrical with respect to the y-axis.
Practice Test
The views are different because the period of the wave is 125. Over a bigger domain, there will be more cycles of the graph.