Trig functions - graphing TYS 1 Solns

Given an equation, describe the features.	1. The main features for the curve y = 3cos 2x are: the amplitude is 3; the number of patterns in 2π is 2 - so the period of the cos function is π.
	2. The main features for the curve y = 2 + sin x are. the amplitude is 1; the graph is shifted up 2 units - so its maximum value is 3 and its minimum value is 1 the number of patterns in 2π is 1 - so the period of the sin function is 2π.
	3. The main features for the curve y = 2 - 3 cos 4x are: the amplitude is 3; the graph is shifted up 2 units - so its maximum value is 5 and its minimum value is -1; the negative in front of the cos term flips the graph vertically about y = 2; the number of patterns in 2π is 4 - so the period of the cos function is π/2.
	4. The main features for the curve are: the amplitude is still infinite; the y values are twice what they were originally - so the curve is stretched up and down in the vertical direction; the number of patterns in π (for tan!!) is ½ - so the period of the tan function is now 2π.
	5. The main features for the curve are: the amplitude is 2; the graph is shifted up 1 unit - so its maximum value is 3 and its minimum value is -1; the negative in front of the sin term flips the graph vertically about y = 1; the number of patterns in 2π is 1 - so the period of this sin function is 2π; there is a phase shift (horizontal movement) of π/4 to the left.
Given a graph, interpret the features.	6. In the graph above of y = a cos nx. a is the amplitude - so a = 2; n is the number of patterns in 2π. Here there is half a pattern in 2 radians giving 1 pattern in 4 radians. - so .
	7. The equation of the following graph is written in the form y = A + B cos(Cx - D). A is the vertical shift - so A = 4; B is the amplitude - so B = (6 - 2)÷2 = 2; C is the number of patterns in 2π. As there is one pattern in 6 and a bit x units (so 2π) C = 1; D is the phase shift (horizontal movement) which is zero.
	8. The equation of the following graph is written in the form y = A + B tan(Cx + D). A is the vertical shift - there is none (one curve passes through 0) so A = 0; B is the "vertical stretch factor". After establishing C = 2, at π/8, y = 3. So B = 3. C is the number of patterns in π. Here there are two patterns so C = 2; D is the phase shift (horizontal movement) which is zero.
	9. The equation of the following graph is written in the form y = A + B sin(Cx + D). A is the vertical shift - so A = 3 (check where the middle is); B is the amplitude - so B = (5 - 1)÷2 = 2. By moving the curve to the left (see value for D) the shape of the curve is just that for a normal sine curve. C is the number of patterns in 2π. There is one pattern from -π/6 to 11π/6 units (so 2π) - hence C = 1; D is the phase shift (horizontal movement from the original position) which is π/3 to the right (NOTE: y = 3 is where the x-axis would have been in the original sin curve. So the curve has been moved (i.e translated) along the line y = 3 by π/3 to the right). So D = - π/3.
	10. The graph of D = A + Bcos Ct is given below. (i) The amplitude is (-1+5)÷2 = 2; there is one pattern in 2 units - so period = 2. (ii) the cosine curve is moved down - so A = -3; B = 2 (from above); Period = 2 = 2π/n giving n = π. Hence C = π.
	11. The shape is clearly a tan graph. There are two patterns in the interval π - so there must be a 2x term. The horizontal shift is π/4 but in the context of the double pattern. So the shift is really π/2 (which is sort of "halved"). Hence the equation must be .
Sketch graphs with transformations.	12. Sketch y = cos t + 2 for 0 ≤ t ≤ 2π.
	13. For the function y = 2 cos x (i) the amplitude = 2. (ii) the period is 2π. (iii)
	14. For the function y = -2 sin 3x; (i) the period is 2π/3. (ii) .
	15.
	16.
	17.
	18. Note: The vertical asymptotes are at x = -0.75, -0.25, 0.25 and 0.75 as the period is 0.5.
	19.
	20. (i) (ii) Use your graph to solve 1 - 2cos x = 0 in the given domain. When the curve crosses the x-axis (at y = 0), x = ± π/3.
	21. (i) To add the ordinates (y values): At x = 0, the y values are 0 and 1 - so y = 1. At x = x = 3π/4, the y values are ±1/(√2) - so y = 0. etc, etc.
Graphs of reciprocal functions.	22.
	23.
	24.
	25.
	26.
	27.
Given features, draw the graph.	28. Draw a sine graph which has a maximum value of 7 and a minimum value of 1 and has two patterns in its domain of 0 ≤x ≤ 2π.
	29. Draw a tan graph using the domain 0 ≤ x ≤ π/2 with the function having consecutive values of 0 at x = 0 and at x = π/4.
	30. Draw a sec graph which has minimum values of 2 at t = 0 and at t = 2π and a maximum value of -2 at t = π. Use the domain of -π ≤ t ≤ π.
	31. Draw a cosine graph having a minimum value of -4 at x = 0, a maximum value of 0 and having two patterns in its domain of 0 ≤x ≤ 2π.
Determining the number of solutions.	32. (iii) There are two solutions of the equation 2 cos2x = 1 - x. The negative solution is marked N while the positive solutions is just below π/4. (actually at x = 0.714 but you do not need to know that ).
	33. (i) Sketch the graphs of y = 3 cos 2θ and y = 1 for 0 ≤ θ ≤ π. (ii) Use your graphs to obtain two approximate solutions to the equation. The approximate solutions are: just below π/4 = 0.785 - say about one fifth down (0.15). So about 0.63; just above 3π/4 = 2.356 - say about the same distance up as part 1 was down (0.15). So about 2.5. (iii) Solve the equation 3 cos 2θ = 1 using the normal technique for 0 ≤ θ ≤ π (answer to 3 decimal places) and compare your answers with those in part (ii). The two solutions are x = 0.615 and x = 2.526.
	34. (i) Sketch y = tan πx and y = 1 - x on the same diagram for 0 ≤ x ≤ 2. (iii) The number of solutions to the equation tan πx = 1 - x for 0 ≤ x ≤ 2 is 3.
	35. (i) Sketch the graph of y = sin 2x for 0 ≤ x ≤ π using relevant sub-intervals. Given the trig term has 2x, a short interval is desirable - and less than π/4. (ii) Hence find the solutions to √2sin 2x > 1 for 0 ≤ x ≤ π.
	36. (i) Draw a neat sketch of the curve y = 3sin 2x and y = 1 - cos x on the same diagram for 0 ≤ x ≤ 2π. (iii) Hence determine the number of solutions the equation 3 sin 2x + cos x = 1 will have in the given domain. The number of solutions is 5.