MAT112 Test #1 Spring '99

1. Consider the angle q = 7p /6 radians.
1. Determine its reference angle: p /6, since q' = q - p in the third quadrant
2. Convert the angle to degree measure: q = 210^o, (also accepted 30^o)
3. Determine the exact value of the following trigonometric functions:
1. tan(q ) = sqrt(3)/3
2. sec(q ) = -2sqrt(3)/3 or -2/sqrt(3)
3. sin(q ) = -1/2

2. Consider the real number 1.4. Use your calculator to determine the following angles accurate to two decimal places:
1. The arctan(1.4) in radians. 0.95
1. What is the range of the arctangent function? (-p/2, p/2)
2. The arcsin(1.4) in degrees. undefined, 1.4 is not in the domain
1. What is the domain of the arcsine function? [-1, 1]
3. The arccos(1.4) in radians. undefined, 1.4 is not in the domain
1. What is the domain of the arccos function? [-1, 1]

3. Find the exact value of the six trigonometric functions of the angle q shown in the figure.
   opp = -5, adj = 2, hyp = sqrt(29)
   sin q = -5/sqrt(29)
   cos q = 2/sqrt(29)
   tan q = -5/2
   csc q = -sqrt(29)/5
   sec q = sqrt(29)/2
   cot q = -2/5

4. Find the exact value of the expression sec[arccot(-11/4)].
   arcot(adj/opp) = q , adj = -11, opp = 4, hyp = sqrt((-11)² + 4²) = sqrt(137)
   sec q = hyp/adj = -sqrt(137)/11
5. Use an inverse trigonometric function(s) to write q as a function of x for each of the triangles.

First triangle	Second triangle
q = arcsin(x/7)	q = arctan(3.5/x) - arctan(1/x)

6. Write an expression that amplifies the cosine function 3 times, has a vertical translation of 18 units up, a period of 2 and a phase shift 1/2 unit left.
   f(x) = d + acos(bx - c).
   Period = 2p /b, b = 2p /2 = p,
   c/b = -1 , c = -1 b = -p /2.
   f(x) = 18 + 3cos(p x + p /2).
7. A rescue airplane flying at an altitude of 5 kilometers sees a ship in distress and a Coast Guard ship directly to the left of the plane. The angles of depression to the ships are 20^o for the ship in distress and 52^o for the rescue vessel. How far apart are the ships?
   d = 5cot(20^o) - 5cot(52^o) = 9.83 km.
   d = 5tan(52^o) - 5tan(20^o) = 4.58 km. is the answer if angle of depression was misinterpreted. See dwg. in key 1a.
8. A ball and spring system at rest is 10 inches long. By pulling down on the ball 5 inches, then releasing the ball, a simple harmonic motion is established. The period for one cycle is measured as 2 seconds. Write a model for this system and determine the balls position (above or below the rest value of 10 inches) after 9 seconds.
   y = a sin(bx), y = acos(bx),
   b = 2p /period = 2p /2 = p .
   a = -5, as the ball is initially displaced in the downward direction,
   y = -5 cos(p x), the choice of sine or cosine is a subtlety based on where the timing starts (at a maximum value use cosine, at a rest value use sine).
   y(9) = -5 cos(9p) = 5, y(9) = -5 sin(9p ) = 0. Either answer accepted, cosine is the correct function.
9. Determine the five key points (max, min, intercepts) for the function
   f(t) = 3 sin(t - p ) (accurate to 2 decimals). Then draw two cycles of this function on the coordinate plane.
   Amplitude = |a| = 3, b = 1, period = 2p , left end point = c/b = p ,
   increment = period/4 = p /2.
   Key Points: (p , 0), (3p /2, 3), (2p , 0), (5p /2, -3), (3p , 0)

10. Consider a central angle q of 120 degrees on a circle of radius 5 centimeters.
1. Find the length of s of the arc that is cut off. s = r q .
   s = 5 (2p /3) = 10p /3 » 10.47 cm
2. Calculate the area A of the sector that is cut off. A = 1 r² q .
   A = 1 (5²) 2p /3 = 25p /3 » 26.18 cm².
3. Determine the area A₀ of the shaded region.
   A_o = A - A_tri = 25p /3 - 6.25sqrt(3) » 15.35 cm².

b = 5 sin 60^o = 5 sqrt(3)/2 = 5sqrt(3)/2, h = 5 cos 60^o = 5 (1 ) = 2.5.

A_tri = 2(1 bh) = 2 ( 1 (5sqrt(3)/2 (2.5))) = 6.25sqrt(3)

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by Jack Tompkins