Download Math 1405 Trigonometry — Undefined Value

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20 Russ Thur 3:10pm | 21 Bryan Fri 9:00am | 22 Bryan Fri 12:00pm | 23 Russ Fri 3:10pm
Math 1405 Trigonometry — Undefined Value
Undefined Value, Spring 2017
Instructions:
• Show all your work, using the space provided on the exam. Don’t skip steps. Use
correct mathematical notation. Write your explanation in words if you need to convey
how you arrived at your answer.
• Write legibly. Failure to write legibly will reduce your probability of earning partial
credit.
• Scratch paper is available from the proctor, if needed. Just ask for it.
• Provide an exact answer for each problem, in simplest form. In problems with computation, clearly mark your final answer by circling it.
• Use of any other items, including the use of a phone, is strictly prohibited. All phones
should be turned off and brought to the front of the room.
• Present your Photo I.D. when turning in your exam.
• The exam has 5 pages. Please check to see that your copy has all the pages.
• Mark your Discussion Section above.
• Failure to follow these instructions will cause a reduction of points in your score.
5
1. Suppose that sin θ = − 12
and
3π
2
< θ < 2π. Find the exact values of cos 2θ .
Solution:
12
θ
b
5
52 + b2 = 122 , so b =
√
√
119
because
119. Then cos θ = − 12
cos = ±
θ
2
√
+ cos θ) = ±
√
√
1
= ± 24
(12 + 119)
√
√
12
+
119
= − 21
6
The answer is negative because
1
(1
2
3π
4
<
θ
2
< π.
3π
2
√
1
(1
2
< θ < 2π. Then
√
+
119
12
Page 2
2. Simplify
csc(−x) − 1
to an expression with only one trig function.
1 − sin(−x)
Solution:
csc(−x) − 1
− csc x − 1
=
1 − sin(−x)
1 + sin x
1
+1
= − sin x
1 + sin x
=−
by the even/odd identities
1+sin x
sin x
1 + sin x
1 + sin x
=−
sin x(1 + sin x)
1
=−
sin x
= − csc x
by a reciprocal identity
1
+
3. Using algebra and identities you needed to memorize, simplify the expression
sec x + 1
1
down to a constant.
cos x + 1
Solution:
1
1
cos x + 1 + sec x + 1
+
=
sec x + 1 cos x + 1
(sec x + 1)(cos x + 1)
cos x + 1 + sec x + 1
=
sec x cos x + cos x + sec x + 1
cos x + 1 + sec x + 1
=
1 + cos x + sec x + 1
=1
by distributing
by a reciprocal identity
Math 1405 Trigonometry — Undefined Value
Page 3
4. Find all the solutions of the equation sin x + cos 2x = 1 in the interval [0, 2π).
Solution: Since sin x + cos 2x = sin x + 1 − 2 sin2 x = −2 sin2 x + sin x + 1, we are
solving the equation
0 = −2 sin2 x + sin x
= − sin x(2 sin x − 1)
sin x = 0, 1/2
x = 0, π, π/6, 5π/6
5. Find all solutions of the equation sec x = 1 − tan x in the interval [0, 2π)
Solution:
(sec x)2 = (1 − tan x)2
sec2 x = 1 − 2 tan x + tan2 x
sec2 x = sec2 x − 2 tan x
0 = −2 tan x
0 = tan x
x = 0, π
However sec(π) =
̸ 1 − tan(π), so π is not a solution. Therefore the only solution to
this equation is x = 0.
Page 4
6. Find all solutions to the equation 6 sin2 x + 5 sin x − 4 = 0.
Solution:
(2 sin x − 1)(3 sin x + 4) = 0
sin x = 1/2, −4/3
sin x ̸= −4/3, because −4/3 is outside the range of sin x, so we only need to solve
sin x = 1/2, whose solutions are x = π/6 + 2πk, 5π/6 + 2πk, k ∈ Z.
√
7. Find all solutions of the equation 3 3 cot(3x) + 3 = 0 in the interval [0, 2π).
Solution: We need to solve cot(3x) = − 3√3 3 , which is equivalent to tan(3x) =
√
3. This is true when 3x = π/3 + πk, k ∈ Z. And so x = π/9 + kπ/3 =
π/9, 4π/9, 7π/9, 10π/9, 13π/9, 16π/9 in [0, 2π).
Math 1405 Trigonometry — Undefined Value
Page 5
8. Simplify the expression 2 cos2 (3θ) − 1
Solution: This follows the formula cos(2x) = 2 cos2 x − 1, and so 2 cos2 (3θ) − 1 =
cos(2 · 3θ) = cos(6θ). That is the best you can do for this problem.
9. Graph the function y = − 23 tan( 32 x) either by using individual transformations (on separate draft graphs) or by working on the asymptotic intervals. Only your final graph,
should be on the axes provided. Label each asymptote with its defining equation and
each point, of which you need at least one, with their ordered pairs. Draw the function
to the ends of the axes. Graphing the function by plotting points will receive no credit.
Finally, state the period.
Solution: We need to solve − π2 < 23 x < pi 2 to determine the principal interval. This
is − π3 < x π3 The vertical scale (a compression and a flip) is shown below on the
graph. This is manifested by the fact that I plotted the point equivalent to ( π4 , 1) on
the parent graph.
3
2
1
− 3π
2
4π 5π 7π
3
4
6
−π
5π 3π 2π
6
4
3
−π
2
π
3
π
4
π
6
π
6
−1
π
4
π
3
π
2
2π 3π 5π
3
4
6
π
( π6 , − 23 )
−2
x = −π
x=
−π
3
−3
x=
π
3
x=π
7π 5π 4π
6
4
3
3π
2