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Exam #2 Solutions · Monday November 20, 2006
MATH 111 · Section 7 · Fall 2006 Name
The solutions I’ve given here include my explanations and thus are wordier than I would expect your responses to be. Only problem 13 is a writing problem; for the rest, computational answers suffice.
Problem 1. Given the function y = 13 sin(4x + 1), find the average value and period.
Solution: The average value is what’s added to the sin term, which is zero. For the period, solve the inequality
0 ≤ 4 x + 1 ≤ 2 π − 1 ≤ 4 x ≤ 2 π − 1
−
≤ x ≤
2 π 4
The period is the right-hand side minus the left-hand side, which is π/2.
Problem 2. Given the following graph, find the maximum value, minimum value, and amplitude.
0 1 2 3 4 5 6 7 8 9 10
−
−
0
1
2
3
4
5
6
Solution: Read off the values max = 5, min = -1, and amplitude = 3. Or, amplitude is (max − min)/2 = (5 + 1)/2 = 3.
Problem 3. Write an equation for a sine function which passes through all the points in the following data table: x 0 1 2 3 4 5 y − 1 3 7 3 − 1 3
Solution: First graph the function:
0 1 2 3 4 5
−
−
0
1
2
3
4
5
6
7
8
Now read off the amplitude, average value, period, and phase shift from the graph and write down
y = avg. val. + amplitude · sin
2 π period
(x − phase shift)
y = 3 + 4 sin
2 π 4
(x − 1)
= 3 + 4 sin
( (^) π 2
(x − 1)
Problem 4. Which of the following correctly represents cos(−θ) + sin(−θ)?
(A) cos(θ) + sin(θ) (B) cos(θ) − sin(θ) (C) − cos(θ) + sin(θ) (D) − cos(θ) − sin(θ) (E) None of these.
Solution: Since cosine is an even function and sine is an odd function, we have, for all θ, cos(−θ) = cos(θ) and sin(−θ) = − sin(θ). So, the answer is B.
Problem 5. Verify the following trigonometric identity. (Hint: use a sum identity and the Pythagorean identity.) cos(2x) = 1 − 2 sin^2 (x).
Solution: The sum identity for cosine gives us
cos(2x) = cos(x + x) = cos(x) cos(x) − sin(x) sin(x) = cos^2 (x) − sin^2 (x).
You’re asked to get the right-hand side looking like 1 − 2 sin^2 (x), but there is a cos^2 (x) term in there to be gotten rid of. Fortunately, we know from the Pythagorean identity that
cos^2 (x) + sin^2 (x) = 1.
Using this, we can put cos^2 (x) = 1 − sin^2 (x)
into the above, picking up from where we left off, to obtain
cos(2x) = cos^2 (x) − sin^2 (x) = (1 − sin^2 (x)) − sin^2 (x) = 1 − 2 sin^2 (x).
Problem 6. Determine the exact value of sin(7π/12) using sum and/or difference identities.
Solution: This is perhaps easier in degrees, perhaps not. Using degrees, we have
7 π 12
7 π 180 ◦ 12 π
= 7 · 15 ◦^ = 105◦.
Now, 105◦^ = 60◦^ + 45◦^ so we can take advantage of known values for sine and cosine at 60◦^ and 45◦, along with the sum identity for sine:
sin(60◦^ + 45◦) = sin(60◦) cos(45◦) + cos(60◦) sin(45◦)
=
Using radians, we have 7 π 12
4 π 12
3 π 12
π 3
π 4 so we can take advantage of known values for sine and cosine at π/3 and π/4, along with the sum identity for sine:
sin(π/3 + π/4) = sin(π/3) cos(π/4) + cos(π/3) sin(π/4)
=
Remember that switching between radians and degrees changes the units of the input to the trigono- metric functions, but doesn’t change the output. So, we get the same answer either way.
Problem 8. Which of the following is not an identity?
(A) sin^2 (x) + cos^2 (x) = 1 (B) cos(x + y) = cos(x) cos(y) + sin(x) sin(y) (C) sin(2x) = 2 sin(x) cos(x) (D) cos(π/ 2 − x) = sin(x) (E) These are all identities.
Solution: B is not an identity. The correct version of the sum identity for cosine is
cos(x + y) = cos(x) cos(y) − sin(x) sin(y).
Answer A is the Pythagorean identity; C is the double-angle identity for sine, which follows immediately from the sum identity for sine; D is a cofunction identity.
Problem 9. Find all solutions for 0 ≤ x < 2 π (or, if you prefer, do it in degrees for 0◦^ ≤ x < 360 ◦):
(sin(x) + 1/2)(sin(x) − 1 /2) = 0.
Solution: Whenever the product of two (or more) real numbers is zero, at least one of them is zero. So, we have two cases:
- If the right-hand factor is zero, then sin(x) = 1/2. There are two points on the unit circle with height 1/2: they are 30◦^ and 150◦.
- If the left-hand factor is zero, then sin(x) = − 1 /2. There are two points on the unit circle with height − 1 /2: they are 210◦^ and 330◦. (If you’re like me, you’d more naturally think of 330◦^ as − 30 ◦. But, the problem asked for solutions between 0◦^ and 360◦.)
So, we have four solutions: 30◦, 150◦, 210◦, and 330◦. In radians, these are π/6, 5π/6, 7π/6, and 11π/6.
Problem 10. Solve for x: y = 5 cos−^1 (2x) + 3.
Solution:
y = 5 cos−^1 (2x) + 3 y − 3 = 5 cos−^1 (2x) y − 3 5
= cos−^1 (2x)
cos
y − 3 5
= 2 x
1 2
cos
y − 3 5
= x.
Problem 11. Find the exact value of cos(sin−^1 (− 2 /3)).
Solution: Remember that trig functions take angles to numbers, while their inverse functions take numbers to angles. So, sin−^1 (− 2 /3) is an angle. Which angle is it? The sin−^1 function has a range of − 90 ◦^ to 90◦; − 2 /3 is negative so we are looking for an angle θ with negative height, where sin(θ) = − 2 /3. We can draw a cartoon to help us get the math right:
6
Z Z Z Z r = 1 ZZ
y = − (^23)
x =?
On the unit circle, r = 1 and so cos(θ) = x/r = x/1 = x. We can use the Pythagorean theorem, x^2 + y^2 = 1, to find
x^2 +
x^2 +
x^2 = 1 −
x = ±
All angles given back by the sine function are between − 90 ◦^ and 90◦^ and so are in quadrants I and IV,
where x is non-negative. So, we choose the positive square root to get x =
√ 5
Problem 12. Find the exact value of arcsin(sin(135◦)).
Solution: From chapter 1, sin(135◦) is one of our well-known values, namely,
2 /2. Now remember that trig functions map angles to numbers, while their inverse functions map numbers to angles. There are, of course, many angles whose sine is
2 /2, and functions (in order to be functions) can have only one output value. We define the inverse sine function to map numbers from −1 to 1 to angles between − 90 ◦^ and 90◦. So, arcsin(
Problem 13. Writing question: Why is it that tan−^1 (2) is defined, while sin−^1 (2) and cos−^1 (2) are undefined?
Solution: The domain of the tangent function is (−∞, +∞) while the domain of the sine and cosine functions is [− 1 , 1]; 2 is in the former but not the latter.
The above sentence is an acceptable answer, of acceptable length. For more explanation of why that, in turn, is true, though:
- Remember that trig functions map angles to numbers, and their inverse functions map numbers to angles.
