=−1.5708 + 0.62236i arcsin
(−1.2 −10i10
)
=−1.5708 − 0.62236i
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.
Compute > Evaluate
arcsinh (sinh (3 + 25i)) = 3− 8iπ + 25i
With the default setting, Solve > Exact nds complex as well as real solutions to trigonometric equations.
Compute > Solve > Exact
tan2x− cot2x = 1 To obtain the principal solutions only1. Choose Tools > Preferences > Computation > Engine.
2. Check Principal Value Only.
Compute > Solve > Exact
tan2x− cot2x = 1, Solution: arcsin
2. At Metropolis Airport, an airplane is required to be at an al-titude of at least 800 ft above ground when it has attained a
horizontal distance of 1 mi 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 (cost, sint) and (t, sint).
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 (cost, sint), (cost,t), and (t, cost), 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
θ
360=2xπ, where x represents the angle in radians. Choose Com-pute > Solve > Exact or ComCom-pute > Solve > Numeric and nameθ as the variable. Use this method to convert x = 60013π radians to degrees.
7. To sol e a triangle means to determine the lengths of the three sides and the measures (in degrees or radians) of the three an-gles.
a. Solve the right triangle with one side of length c = 2 and one angleα =π9.
b. Solve the right triangle with two sides a = 19 and c = 23.
8. e law of sines a
sinα = b
sinβ = c sinγ
enables you to solve a triangle if you are given one side and two angles, or if you are given two sides and an angle opposite one of these sides. Solve the triangle with one side c = 2 and two anglesα =π9,β =2π9.
9. Using both the law of sines and the law of cosines, a2+ b2− 2abcosγ = 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 in-cluded angleγ = 21629π.
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−π2. Find the general solu-tion.
Solutions
1. De ning functions f (x) = x3+ x sin xand 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 consider-ing the right triangle with legs of length 800 ft and 5280 ft. e angle in question is the acute angle with sine equal to√ 800
8002+52802. Find the answer in radians with Compute > Evaluate > Nu-meric:
arcsin 800
√8002+ 52802 ≈ 0.15037
You can express this angle in degrees by using the following steps:
360×0.15037
2π ≈ 8.6157 0.6157× 60 = 36.942 θ ≈ 8◦37′
or solve the equation 0 .15037 rad = x◦to get 8. 615 6, then solve 0.6156◦= x′to get 36. 936.
3. Note that We leave the conjecture up to you.
4. Figure 4a shows a circle of radius 1 with center at the origin. e graph is drawn by starting at the point (1, 0) and is traced in a counter-clockwise direction. Figure 4b shows the y-coordinates of the rst gure as the angle varies from 0 to 2π. e point (cos 1, sin 1)is marked with a small circle in the rst gure. e corresponding point (1, sin 1) is marked with a small circle in the second gure. 5. Figure 5a shows a circle of radius 1 with center at the origin. e
graph is drawn by starting at the point (1, 0) and is traced in a counter-clockwise direction.
Figure 5b shows the x-coordinates of the rst gure as the an-gle varies from 0 to 2π. e point (cos1,sin1) is marked with a small circle in the rst gure. e corresponding point (cos 1, 1) is marked with a small circle in the second gure.
Figure 5c shows the graph in the second gure with the hori-zontal and vertical axes interchanged. Figure 5c shows the usual view of y = cos x. equation, choose Compute > Solve > Exact to getθ =3910
de-grees, or choose Numeric to getθ = 3.9 degrees.
7. To obtain the solutions in the simple form shown below, choose Compute > Engine Settings and check Principal Value Only.
a. Choose Compute > De nitions > New De nition for each of the given valuesα =π9 and c = 2. Evaluateβ =
π2−α toget β = 187π. Evaluate csinα toget a = 2sin19π (= 0.68404). Evaluate c cosα to get b = 2cos19π (=
1.8794).
b. Choose Compute > De nitions > New De nition to each of the given values, a = 19 and c = 23. Place the in-sert point in the equation a2+ b2= c2and choose Com-pute > Solve > Exact (Numeric) to get b = 2√
42(=
12.96). Place the insert point in each of the equations sinα = a
c, cosβ = a
c in turn, and choose Compute >
Solve > Exact to getα = arcsin1923,β = arccos1923; or place the insert point in each of the one-column matrices [ sinα = a/c
in turn, and choose Compute >Solve > Numeric to getα = 0.9721, β = 0.5987. 8. Choose Compute > De nitions > New De nition to de ne
α = π9,β = 2π9 , and c = 2. Evaluateγ = π − α − β to get γ = 23π. De ne γ = 23π. Choose Compute > Solve > Exact to solve the equations a
sinα = c
3 sin29π. To get numerical solutions, choose Compute > Solve > Numeric.
9. Solving general triangles.
a. De ne each of a = 2.34, b = 3.57, andγ =21629π. Choose Compute > Solve > Exact to to solve the equation a2+ b2−2abcosγ = c2. You should get c = 1.7255. De ne c = 1.7255. Choose Compute > Solve > Exact to solve the equations a
sinα = c
sinγ and b
sinβ = c
sinγ, or with
the insert point in each of the matrices a
c
choose Compute > Solve > Numeric to getα = 0.58859 andβ = 1.0104.
A triangle with three sides given is solved similarly: inter-change the actions onγ and c in the steps just described.
b. De ne a = 2.53, b = 4.15, and c = 6.19. Choose Com-pute > Solve > Exact to solve the equation a2+ b2− 2ab cosγ = c2to getγ = 2.3458. De ne γ = 2.3458.
Choose Compute > Solve > Exact to solve each of the equations a
sinα = c
sinγ and b
sinβ = c
sinγ, or put the insert point in each of the matrices
and choose Compute > Solve > Numeric to getα = 0.29632 andβ = 0.49948.
10. In polar form,
i = cosπ For the general solution, for any integer k,
i = cos(π