MAT112 Test #1 Spring '99

1. Consider the angle q = 4p/3 radians.
1. Determine its reference angle: p/3, since q' = q -p in the third quadrant
2. Convert the angle to degree measure: q = 240^o, (also accepted 60^o)
3. Determine the exact value of the following trigonometric functions:
1. tan(q) = sqrt(3)
2. sec(q) = -2, since sec(q) = 1/cos(q) = 1/1 = 2, and negative due to third quadrant
3. sin(q) = -sqrt(3) / 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 arccos(1.4) in degrees. undefined, 1.4 is not in the domain
1. What is the domain of the arccosine function? [-1, 1]
3. The arcsec(1.4) in radians. 0.78, since arcsec(1.4) = arccos(1/1.4)
1. What is the range of the arcsecant function? [0, p ], q ¹ p /2

3. Find the exact value of the six trigonometric functions of the angle q shown in the figure.
   opp = -4, adj = 1, hyp = sqrt(17):
   sin q = -4/sqrt(17) = -4 sqrt(17)/17,
   cos q = 1/sqrt(17) = sqrt(17)/17,
   tan q = -4,
   cscq = -sqrt(17)/4,
   sec q = sqrt(17),
   cot q = -1/4.
   
4. Find the exact value of the expression cos[arccot(-11/4)].
   cos q = adj / hyp = -11 / sqrt(137) = -11sqrt(137)/137.
   
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 sine 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 + asin(bx - c).
   Period = 2p /b, b = 2p /2 = p ,
   c/b = -1 , c = -1 b = -p /2.
   f(x) = 18 + 3sin(p x + p /2).

7. A passenger in an airplane flying at an altitude of 5 kilometers sees two towns directly to the left of the plane. The angles of depression to the towns are 23^o and 52^o. How far apart are the towns?
   d = 5cot(23^o) - 5cot(52^o) = 7.87 km.
   d = 5tan(52^o) - 5tan(23^o) = 4.28 km. is the answer if angle of depression was misinterpreted.
8. A ball and spring system at rest is 15 inches long. By pulling down on the ball 2.5 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 15 inches) after 9 seconds.
   y = a sin(bx), y = a cos(bx), b = 2p /period = 2p /2 = p .
   y = -2.5 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) = -2.5 cos(9p ) = 2.5. y(9) = -2.5 sin(9p ) = 0. Either answer accepted, cosine is the correct function.
9. Determine the five key points (accurate to 2 decimals) for the function
   f(t) = 3 cos(t - p ). 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 , 3), (3p /2, 0), (2p , -3), (5p /2, 0), (3p ,3)

10. Consider a central angle q of 120 degrees on a circle of radius 10 centimeters.
1. Find the length of s of the arc that is cut off. s = r q .
   s = 10 (2p /3) = 20p /3 » 20.94 cm
2. Calculate the area A of the sector that is cut off. A = 1 r² q .
   A = 1 (10²) 2p /3 = 100p /3 » 104.72 cm².
3. Determine the area A₀ of the shaded region.
   A_o = A - A_tri = 100p /3 - 25sqrt(3) » 61.42 cm².

b = 10 sin 60^o = 10 sqrt(3)/2 = 5sqrt(3), h = 10 cos 60^o = 10 (1 ) = 5.

A_tri = 2(1 bh) = 2 ( 1 (5sqrt(3) (5))) = 25sqrt(3)

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by Jack Tompkins