MATH 1090 Algebra and Trigonometry
Problem Set 7 - Solutions

2002 Winter

Note

Without using a calculator, find the exact values of all six trigonometric ratios (sine, cosine, tangent, cosecant, secant and cotangent) for the angle whose measure is

13p/3

13p/3 = (12+1)p/3
= 2×2p + p/3
Reference angle for 13p/3
is p/3 (=60°), in the
first quadrant.
Therefore [reference angle diagram]

cos (13p/3) = 1/2	sec (13p/3) = 2
sin (13p/3) = Ö3 / 2	csc (13p/3) = 2Ö3 / 3
tan (13p/3) = Ö3	cot (13p/3) = Ö3 / 3

-p/6

[reference angle diagram] Reference angle for -p/6
is p/6 (=30°), in the
fourth quadrant.
Therefore

cos (-p/6) = Ö3 / 2	sec (-p/6) = 2Ö3 / 3
sin (-p/6) = -1/2	csc (-p/6) = -2
tan (-p/6) = -Ö3 / 3	cot (-p/6) = -Ö3

1003p/4

1003p/4 = (1000+3)p/4
= 250×2p + 3p/4
Reference angle for 1003p/4
is p/4 (=45°), in the
second quadrant.
Therefore [reference angle diagram]

cos (1003p/4) = -Ö2 / 2	sec (1003p/4) = -Ö2
sin (1003p/4) = Ö2 / 2	csc (1003p/4) = Ö2
tan (1003p/4) = -1	cot (1003p/4) = -1

Without using a calculator, find the exact value, in simplified form, of
1. The reference angle for 7p/4 is p/4 (= 45°) in the fourth quadrant.
  Therefore
  
  Therefore
2. The reference angle for 7p/6 is p/6 (= 30°) in the third quadrant.
  Therefore
  
  Therefore

Without using a calculator, find the exact values of the other five trigonometric ratios, given that
1. sin x = -3/5 and x is in the fourth quadrant.
  
  x is in the fourth quadrant Þ cos x and sec x are positive, but the other four trigonometric ratios are negative.
  $sin x = -3/5 ==> cos x = +sqrt{1 - sin^2 x} = sqrt{1 - 9/25} = sqrt{16/25} = 4/5$
  
  OR read the values from this right-angled triangle:
2. tan t = 12/5 and p < t < 3p/2.
  
  t is in the third quadrant Þ tan t and cot t are positive, but the other four trigonometric ratios are negative.
  
  $tan t = 12/5 ==> sec t = -sqrt{1 + tan^2 t} = -sqrt{1 + 144/25} = -sqrt{169/25} = -13/5$
  
  $sin t = -sqrt{1 - cos^2 t} = -sqrt{1 - 25/169} = -sqrt{144/169} = -12/13$
  
  OR read the values from this right-angled triangle:

[Challenge Question:]
Given that the exact value of sin 15° is find the exact values of
1. cos 75°
  
  Using the properties of cofunctions and complementary angles,
  cos 75° = sin (90-75)°
  Therefore
2. sin 75°
  
  $sin 75deg = sqrt{1-cos^2 75deg} = sqrt{1 - (2 - sqrt{3})/4} = sqrt{(4 - 2 + sqrt{3})/4}$
  Therefore
3. $= sqrt{(2 + sqrt{3})^2 / (4 - 3)}$
  Therefore

Sketch the graphs of
1. y = sin 2x
  Comparing y = sin 2x to the general form y = A sin (Bx-C) we have
  Amplitude A = 1 , frequency B = 2 and phase shift -C/B = 0
  Consecutive x intercepts occur p/B = p/2 units apart.
  The graph is identical to y = sin x except for a horizontal compression by a factor of 2.
  The sketch then follows.
2. y = cos (t - p/4)
  This is clearly y = cos t after a [phase] shift to the right of p/4.
  The first peak at (t, y) = (0, 1) is shifted to (p/4, 1).
  The right-most negative zero is shifted from (-p/2, 0) to (-p/4, 0).
  The period remains 2p.
3. y = cot^-1 x
  The graph of y = cot^-1 x is the reflection in the line y = x of part of the graph of y = cot x , a standard form that appears on page 502 of the course textbook.
  Restrict the domain to [0, p], so that the function will pass the horizontal line test for an inverse function to exist.
  The graph then follows immediately.

Where possible, evaluate the following without using a calculator.
1. tan^-11
  tan p/4 = 1 and p/4 is in the range of the inverse tangent function
  Þ
2. cos (cos^-12)
  2 is not in the domain of the inverse cosine function
  Þ
3. sin^-1 (sin (7p/6))
  sin 7p/6 = sin -p/6 = -1/2 and -p/6 is in the range of the inverse sine function;
  but 7p/6 is not in the range of the inverse sine function.
  Þ

Also: Try the questions from exercise sets 5.1 to 5.7 of the textbook.

[Index of Questions]

[Index of Solutions]

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Dr. G.H. George

MATH 1090 Algebra and Trigonometry Problem Set 7 - Solutions

MATH 1090 Algebra and Trigonometry
Problem Set 7 - Solutions