Math140 Graph of Tangent Function

Graph of Tangent Function x y = tan x y=sin x / cos x -π/2 =-1/0=UNDEFINED -π/3 -1.732 -π/4 -1.000 -π/6 -0.577 0 0.000 π/6 0.577 π/4 1.000 π/3 1.732 π/2 (− (− ) 3 2 (− 1 2) = − 3 2 2 )( ( 3 2) = -1 (−1 2) ) 2 2 = −1 3 =− 3 3 = 0/1 = 0 ) 3= 3 ( 2 2 ) ( 2 2 )= 1 ( 3 2 ) (1 2 ) = 3 (1 2 ) ( 3 2 =1 =1/0= UNDEFINED 2π/3 -1.732 3π/4 -1.000 5π/6 -0.577 π 0.000 7π/6 0.577 5π/4 1.000 4π/3 1.732 3π/2 =0/-1 = 0 = -1/0 = UNDEFINED 5π/3 -1.732 7π/4 -1.000 11π/6 -0.577 2π 0.000 13π/6 0.577 9π/4 1.000 7π/3 1.732 = 0/1 = 0 3 Recall that the domain of the tangent function is all real numbers except for odd multiples of π/2 (i.e. π/2, 3π/2, 5π/2, etc.), because at those values the tangent function is undefined. The vertical lines through these x-values are the vertical asymptotes of the graph. We should also note that as x approaches values close to an odd multiple of π/2, the absolute value of y = tan x gets very large. Notice that π/2 ≈ 1.57. If we let x= 1.56, y= tan 1.56 ≈ 92.62. We also know that the period (or cycle) of the tangent function is π, which means the values of the tangent function repeat themselves after the length of π on the x-axis. We can see from the graph that the tangent function is symmetric with respect to the origin, so the tangent function is odd. This means tan(-x) = - tan x. Vertical Asymptotes at - π/2, π/2,3π/2, 5 π/2, etc. 2 5π/2 1 Where are the xintercepts? (9π/4, 1) 0 π/2 -π/2 Where are the yintercepts? (5π/4, 1) (π/4, 1) (-π/4, -1) -1 -2 (3π/4, -1) π 3π/2 2π (7π/4, -1) 5π/2 Transformations of Tan x Example 1 Graphing Variations of y=tan x using tranformations. Graph y = 2tan x 2 is our amplitude, A, which means it is the factor that will vertically stretch or shrink our graph. Since A=2, our graph will vertically stretch by a factor of 2. For instance, when x =π/4, tan x = sin π/4/cos π/4 = 1, so 2tan π/4 = 2(1) = 2 3 2 (5π/4, 2) (π/4, 2) (9π/4, 2) 1 0 π/2 -π/2 π 2π 3π/2 5π/2 -1 (-π/4, -2) (3π/4, -2) -2 (7π/4, -2) -3 π Example 2 Graph y = − tan⎛⎜ x + ⎞⎟ ⎝ 4⎠ This graph will have 2 changes. First, since it is negative, it be reflected on the x-axis (turned upside-down). Also, as in our y = Asin(wx+h) + v, h is the horizontal displacement. Since h >0, the graph is shifted to the left. y = -tan x y = -tan (x+ pi/4) 2 2 1 1 0 π/2 -π/2 0 π 3π/2 2π 5π/2 -π/2 -1 -1 -2 -2 π/2 π 3π/2 2π 5π/2 Vertical Asymptotes now at π/4, 5π/4 ,9π/4, etc. Graph of Cot Function Recall that the domain of the cotangent function is all real numbers except for all integral multiples of π (i.e. 0, π, 2π, 3π, etc.), because at those values the tangent function is undefined (at those values cos x = ±1 and sin x = 0, and cot x = cos x/ sin x). Vertical Asymptotes 2 -π/2 (-π/4, -1) 1 (π/4, 1) 0 π/2 -1 -2 (9π/4, 1) (5π/4, 1) π (3π/4, -1) 3π/2 2π (7π/4, -1) 5π/2 2.8 Phase Shifts and Sinusoidal Curve Fitting Previously, when talking about transformations, I used the following equation for a sinusoidal graph: y = Asin(ωx + h) + v This is not the standard form. The standard form is: y = Asin(ωx - φ) + B Notice in this form we use – φ, instead of + h. So this means if φ > 0, the graph is shifted to the RIGHT, and if φ < 0, the graph is shifted to the LEFT. In the previous form, I said that h was the horizontal displacement. However, the actual amount of the horizontal displacement, or phase shift, is φ/ω. How do we get this? Recall that the period of the sine function is 2π. sin(ωx - φ) = sin((ωx – φ) + 2π ) The period will begin with ωx – φ = 0 Æ x = φ/ω = starting point of one cycle. We see from this that this shows that the starting point is shifted from 0 to φ/ω. And end with ωx – φ = 2π Æ x = 2π/ω + φ/ω = ending point of one cycle. Also remember that the new period is T = 2π/ω Example 1 Find amplitude, period, phase shift of y = 3sin(2x –π) A = 3, ω = 2, φ = π Amplitude = A = 3 This that the y values range from -3 to 3. Period T = 2π/ω = 2π/2 = π This means that the whole sine cycle is squeezed into a period of π. Phase shift = φ/ω = π/2. x = φ/ω = starting point of one cycle = π/2. x = 2π/ω + φ/ω = ending point of one cycle = 2π/2 + π /2 = 3π/2 Since φ > 0, the graph is shifted to the right, so start the cycle at x = φ/ω = π/2. 4.000 3.000 2.000 1.000 0.000 -1.000 -2.000 -3.000 -4.000 π/4 π/2 3π/4 π 5π/4 5π/4 3π/2 2π Example 3 Finding a Sinusoidal Functions from Temperature Data Average Monhly Temp Month 1 29.7 2 33.4 3 39 4 48.2 5 57.2 6 66.9 7 73.5 8 71.4 9 62.3 10 51.4 11 39 12 31 Average Montly Temperature for Jan-Dec. 80 70 60 50 40 Series1 30 20 10 0 0 1 2 3 4 5 6 7 8 9 10 11 12 We will fit this to the STANDARD sinusoidal function: y = Asin(ωx - φ) + B Step 1: To find the amplitude A, we compute A = largest data value - smallest data value = 2 73.5 − 29.7 = 21.9 2 Step 2: Determine the vertical shift, B, by finding the average of the highest and lowest value. B = 73.5 + 29.7 2 = 51.6 Step 3: Find frequency, ω. Cycle will (hopefully) repeat itself every year, so period = 12 months. T = 12. Use the fact that T = 2π/ ω to solve for ω. ω = 2π/T = 2π/12 = π/6 Step 4: Using A=21.9, ω = π/6, and B = 51.6,determine horizontal shift by choosing an arbitrary data point (x,y) from the given table and solving the equation for φ. ⎛π ⎞ y = 21.9 sin ⎜ x − ϕ ⎟ + 51.6 ⎝6 ⎠ Let’s choose the first data point, x = 1 (Jan.), y = 29.7 ⎛π ⎞ 29 .7 = 21 .9 sin ⎜ (1) − ϕ ⎟ + 51 .6 ⎝6 ⎠ ⎛π ⎞ − 21 .9 = 21 .9 sin ⎜ − ϕ ⎟ ⎝6 ⎠ ⎛π ⎞ − 1 = sin ⎜ − ϕ ⎟ ⎝6 ⎠ When does sin = - 1? For consistenc y, we will choose an angle between π -π and . 2 2 ⎛ -π ⎞ sin ⎜ ⎟ = −1 ⎝ 2 ⎠ π 6 π 6 −ϕ = − + π 2 π 2 =ϕ = Solve for ϕ by getting all other term s on the other side. π 6 + 3π 4π 2π = = 6 6 3 So.. A sine function that fits the data is 2π ⎛π y = 21.9 sin ⎜ x − 3 ⎝6 ⎞ ⎟ + 51.6 ⎠ Avg Monthly Temp y=21.9*sin((pi/6)*x - 2pi/3) + 51.6 80 70 60 50 40 30 20 0 1 2 3 4 5 6 7 8 9 10 11 12 13