All trigonometric, inverse trigonometric, and hyperbolic functions are de ned for com-plex arguments. Arguments that are rational multiples of i are rewritten in terms of hyperbolic functions.
The function arcsinh produces values with imaginary parts in the interval 2;2 . I Evaluate
sin 5i = i sinh 5 cos4i5 = cosh54 tan ( 3i) = i tanh 3 arcsin 5i = i arcsinh 5 arccos4i5 =12 + i arcsinh54 arctan ( 3i) = i arctanh 3
Hyperbolic functions with arguments that are integer multiples of 2iare simpli ed by Evaluate.
I Evaluate
sinh 2i = i cosh (40i ) = 1 tanh 10100i = 0 arccosh 0 = 12i arccoth 0 = 12i
For other complex arguments, use Expand to rewrite trigonometric and hyperbolic functions.
I Expand
sin 5i + 23 =12p
3 cosh 5 12i sinh 5 cos 4i5 4 = 12p
2 cosh54 12ip 2 sinh54
tan 3i +2 = tanh 3i sinh (x + i ) = sinh x
Use Rewrite to obtain a representation in terms of speci c target functions.
I Rewrite + Sin and Cos
e2ixtan x = cos x1 (sin x) (cos 2x + i sin 2x)
For arcsin and arccos, the branch cuts are the real intervals ( 1; 1) and (1; 1).
For arctan, the branch cuts are the intervals ( 1 i; i] and [i; 1 i) on the imaginary axis. For arcsec and arccsc, the branch cut is the real interval ( 1; 1). For arccot, the branch cut is the interval [i; i] on the imaginary axis. The values jump when the arguments cross a branch cut.
I Evaluate
arcsin ( 1:2) = 1:5708+0:62236i arcsin 1:2 +10i10 = 1:5708+0:62236i arcsin 1:2 10i10 = 1:5708 0:622 36i
Note that arccot is de ned by arccot x = arctan1xalthough arccot does not rewrite itself in terms of arctan. As a consequence of this de nition, the real line crosses the branch cut and arccot has a jump discontinuity at the origin.
I Evaluate
arcsinh (sinh (3 + 25i)) = 3 8i + 25i
With the default real/complex setting, Solve + Exact nds complex as well as real solutions to trigonometric equations.
I Solve + Exact
tan2x cot2x = 1, Solution is: 12 + arctanp pp2
5+1+ k j k 2 Z [ 12 arctanppp2
5+1+ k j k 2 Z [ 12 arctanp pp2
5+1 + k j k 2 Z [ 12 + arctanppp2
5+1+ k j k 2 Z I To obtain the principal solution only
1. From the Tools menu, choose Engine Setup.
2. On the General page, under Solve Options, check Principal Value Only.
I Solve + Exact
2. At Metropolis Airport, an airplane is required to be at an altitude of at least 800 ft above ground when it has attained a horizontal distance of one mile from takeoff.
What must be the (minimum) average angle of ascent?
3. Experiment with expansions of sin nx in terms of sin x and cos x for n = 1, 2, 3, 4, 5, 6 and make a conjecture about the form of the general expansion of sin nx.
4. Experiment with parametric plots of (cos t; sin t) and (t; sin t). Attach the point (cos 1; sin 1) to the rst plot and (1; sin 1) to the second. Explain how the two graphs are related.
5. Experiment with parametric plots of (cos t; sin t), (cos t; t), and (t; cos t), together with the point (cos 1; sin 1) on the rst plot, (cos 1; 1) on the second, and (1; cos 1) on the third. Explain how these plots are related.
6. To convert radians to degrees using ratios, write the equation
360 = x
2 , where x represents the angle in radians. From the Solve submenu, choose Exact or Numeric and name as the Variable to Solve for. Use this method to convert x = 13 radians to degrees. 600
7. To solve a triangle means to determine the lengths of the three sides and the measures (in degrees or radians) of the three angles.
a. Solve the right triangle with one side of length c = 2 and one angle = 9.
enables you to solve a triangle if you are given one side and two angles, or if you aresin given two sides and an angle opposite one of these sides. Solve the triangle with one side c = 2 and two angles = 9, = 29.
9. Using both the law of sines and the law of cosines, a2+ b2 2ab cos = c2
you can solve a triangle given two sides and the included angle, or given three sides.
a. Solve the triangle with sides a = 2:34, b = 3:57, and included angle = 21629 . b. Solve the triangle with three given sides a = 2:53, b = 4:15, and c = 6:19.
10. Fill in the steps to show that ii = e 2. Find the general solution.
Solutions
1. De ning functions f(x) = x3+ x sin x and g(x) = sin x2and evaluating gives f (g(x)) = sin3x2+ sin x2sin sin x2
g(f (x)) = sin x3+ x sin x 2 f (x)g(x) = x3+ x sin x sin x2 f (x) + g(x) = x3+ x sin x + sin x2
2. You can nd the minimum average angle of ascent by considering the right triangle with legs of length 800 ft and 5280 ft.
0 2000 4000
0 500
x y
The angle in question is the acute angle with sine equal to 800
p8002+ 52802. Find the answer in radians with Evaluate Numerically:
arcsin 800
p8002+ 52802 = 0 :15037
You can express this angle in degrees by using the following steps:
360 0:15037
2 = 8:6157
0:6157 60 = 36:942 37
= 8 370 or by solving the equations
0 :15037 rad = , Solution is: f = 8: 6156g 0:6156 = x0, Solution is: 36: 936 3. Note that sin 2x = 2 sin x cos x
sin 3x = 4 sin x cos2x sin x sin 4x = 8 sin x cos3x 4 sin x cos x
sin 5x = 16 sin x cos4x 12 sin x cos2x + sin x sin 6x = 32 sin x cos5x 32 sin x cos3x + 6 sin x cos x
We leave the conjecture up to you.
4. The rst gure shows a circle of radius 1 with center at the origin. The graph is drawn by starting at the point (1; 0) and is traced in a counter-clockwise direction.
The second gure shows the y-coordinates from the rst gure as the angle varies from 0 to 2 . The point (cos 1; sin 1) is marked with a small circle in the rst gure.
The corresponding point (1; sin 1) is marked with a small circle in the second gure.
-1 1
5. The rst gure shows a circle of radius 1 with center at the origin. The graph is drawn by starting at the point (1; 0) and is traced in a counter-clockwise direction.
The second gure shows the x-coordinates of the rst gure as the angle varies from 0 to 2 . The point (cos 1; sin 1) is marked with a small circle in the rst gure. The corresponding point (cos 1; 1) is marked with a small circle in the second gure. The third gure shows the graph from the second gure with the horizontal and vertical axes interchanged. The third gure shows the usual view of y = cos x.
-1 1
2 . With the insertion point in this equation, from the Solve submenu, choose Exact to get = 3910 degrees, or choose Numeric to get
= 3:9 degrees.
7. To obtain the solutions in the simple form shown below, choose Tools + Engine Setup. On the General page, under Solve Options, check Principal Value Only.
a. Choose New De nition from the De nitions menu for each of the given values
= 9and c = 2. Apply Evaluate to = 2 to get = 187 . Apply Evaluate (or Evaluate Numerically) to a = c sin to get a = 2 sin19 (= 0:68404).
Apply Evaluate to b = c cos to get b = 2 cos19 (= 1:8794).
b. Apply De nitions + New De nition to each of the given values, a = 19 and
c = 23. Place the insertion point in the equation a2+ b2 = c2 and, from the Solve submenu, choose Exact (Numeric) to get b = 2p
42 (= 12:96). Place the insertion point in each of the equations sin = a
c, cos = a
c in turn, and choose Solve + Exact to get = arcsin1923; = arccos1923; or place the insertion point in each of the one-column matrices sin = a=c
2 (0; =2) and cos = a=c 2 (0; =2) in turn, and choose Solve + Numeric to get = 0:9721; = 0:5987.
8. Use New De nition on the De nitions submenu to de ne = 9, =29, and c =
3 sin29 . To get numerical solutions, you can apply Solve + Numeric or you can evaluate the preceding solutions numerically.
9. Solving general triangles. Numeric to each of the matrices
0
A triangle with three sides given is solved similarly: interchange the actions on and c in the steps just described.
b. De ne a = 2:53, b = 4:15, and c = 6:19. Apply Solve + Exact to a2+ b2
sin , or apply Solve + Numeric to each of the matrices
solu-tion, i = cos 2 + 2 k + i sin 2 + 2 k = ei(2+2 k) and ii= ei(2+2 k) i = e 2 2 kfor any integer k.