. The absolute value of constant A A (amplitude) increases the total range, and constant D D (vertical displacement) moves the graph vertically. For a function to have an inverse, the function must be unambiguous and pass the horizontal line test. The regular sinusoidal function is not one-to-one unless its domain is restricted in some way. Mathematicians agreed to limit the sinusoidal function to the interval [ − π 2 , π 2 ] [ − π 2 , π 2 ] so that it is one-to-one and has an inverse. Graphs are not symmetric to the y=x. y=x line. They are symmetrical with respect to the y-y axis. The graph is symmetric with respect to the y-axis and there is no amplitude because the limits of the function are |x| | x decrease| Shoot. There seems to be a horizontal asymptote at y=0 y=0. Amplitude: 2 3; 2 3 ; Period: 2π; 2π; Center line: y=0; y=0; Maximum: y= 2 3 y= 2 3 occurs at x=0; x=0; Minimum: y=− 2 3 y=− 2 3 occurs at x=π; x=π; For a period, the graph starts at 0 and ends at 2π 2π Amplitude: none; Period: 4π; 4π; no phase shift; Asymptotes: x=2πk, x=2πk, where k k is an integer. No. The angle that the scale makes with the horizontal is 60 degrees.

It would be reflected by the line y=−1, y=−1 and would become an increasing function. Stress factor: 2; Period: 4 π; π 4 ; Asymptotes: x= 1 4 ( π 2 +πk )+8, where k is an integer x= 1 4 ( π 2 +πk )+8, where k is an integer True. The angle θ 1 θ 1, which corresponds to arccos(−x) arccos(−x), x>0 x>0, is a second quadrant angle with reference angle, θ 2 θ 2 , where θ 2 θ 2 arccosx arccosx, x>0 x>0. Since θ 2 θ 2 is the reference angle for θ 1 θ 1, θ 2 =π− θ 1 θ 2 =π− θ 1 and arccos(−x) arccos(−x) = π−arccosx π−arccosx-. SIN −1 (0.6)=36.87°=0.6435 SIN −1 (0.6)=36.87°=0.6435 radians. The function y=sinx y=sinx is one-to-one on [ − π 2 , π 2 ]; [ − π 2 , π 2 ]; Thus, this interval is the interval of the inverse function of y=sinx, y=sinx, f(x)= sin −1 x. f(x)= sin −1 x. The function y=cosx y=cosx is one-to-one on [ 0.π ]; [ 0,π ]; Thus, this interval is the interval of the inverse function of y=cosx,f(x)= cos −1 x. y=cosx,f(x)= cos −1 x. Amplitude: 3; Period: 4 π; π 4 ; Axis: y=5; y=5; Maximum: y=8 y=8 occurs at x=0.12; x = 0.12; Minimum: y=2 y=2 occurs at x=0.516; x = 0.516; horizontal displacement: −4; – 4; vertical gear ratio 5; A period occurs from x=0x=0 to x= π 4x= π 4.

Amplitude: 1; Period: 12; Phase shift: −6; -6; Central line y=−3 y=−3 A linear function is added to a periodic sinusoidal function. The graph has no amplitude, because if the linear function increases without a link, the combined function h(x)=x+sinx h(x)=x+sinx also increases without a link. The graph is bounded between the graphs y=x+1 y=x+1 and y=x-1 y=x-1 because the sine oscillates between −1 and 1. Center line: y=0; y=0; Amplitude: | A |= 1 2; | A |= 1 2; Period : P = 2π | B | =6π; P= 2π | B | =6π; Phase shift: C B = π C B = π. The sine and cosine functions have the property that f( x+P )=f( x ) f( x+P )=f( x ) for a given P. P. This means that the function values for each unit P P are repeated on the x-axis. Maximum: 1 1 to x= 0 x=0; Minimum: -1 -1 to x= π x=π Amplitude: 1; Period: 2π; 2π; Axis: y=1; y=1; Maximum: y=2 y=2 occurs at x=2.09; x = 2.09; Maximum: y=2 y=2 occurs at t=2.09; t = 2.09; Minimum: y=0 y=0 occurs at t=5.24; t = 5.24; phase shift: − π 3; − π 3 ; Vertical gear ratio: 1; A complete period varies from t=0 t=0 to t=2π t=2π. The views are different, since the period of wave 1 is 25. 1 25.

Over a wider range, there will be more cycles of the chart. Since y=cscx y=cscx is the reciprocal function of y=sinx, y=sinx, you can plot the inverse of the coordinates on the graph of y=sinx y=sinx to get the y-coordinates of y=cscx. y=cscx. The x sections of the graph y=sinx y=sinx are the vertical asymptotes of the graph y=cscx. y=cscx. At the point where the end side of t t intersects the unit circle, you can determine that the sint sint is equal to the y-coordinate of the point. Domain [ −1.1 ]; [ −1.1 ]; Range [ 0.π ] [ 0.π ] Amplitude: 2; Central line: y=−3; y=−3; Period: 4; Equation: f(x)=2sin( π 2 x )−3 f(x)=2sin( π 2 x )−3. Two possibilities: y=4sin( π 5 x− π 5 )+4 y=4sin( π 5 x− π 5 )+4 or y=−4sin( π 5 x+ 4π 5 )+4 y=−4sin( π 5 x+ 4π 5 )+4 About the approximate intervals ( 0,0,4 5,1 ),( 1.6,2.1 ),( 2.6,3.1 ),( 3.7,4.2 ),( 4.7,5.2 ),(5.6,6.28) ( 0.5,1 ),( 1.6,2.1 ),( 2.6,3.1 ),( 3.7,4.2 ),( 4.7,5.2 ),(5.6,6.28). π 6 π 6 is the radiant measurement of an angle between − π 2 − π 2 and π 2 π 2 with a sine of 0.5. Amplitude: 4; Period: 2; Center line: y=0; y=0; Equation: f(x)=−4cos( π( x− π 2 ) ) ) f(x)=−4cos( π( x− π 2 ) ) Amplitude: 1; Timeframe: π; π; Center line: y=0; y=0; Maximum: y=1 y=1 occurs at x=π; x=π; Minimum: y=−1 y=−1 occurs at x= π 2; x= π 2; A complete period is graphically from x=0 x=0 to x=π x=π Responses vary. With the unit circle, you can show that tan( x+π ) = tanx.

tan( x+π )=tanx. Amplitude: 3; Period: 2π; 2π; Axis: y=3; y=3; No asymptotes.