|Trigonometry Demystified, 2nd edition|
|Explanations for Quiz Answers in Chapter 13|
|1. In an arithmetic sequence, the terms increase or decrease at a "steady rate." We always add the same constant (positive, negative, or zero) to each term in order to generate the next one. The difference between any pair of adjacent terms therefore equals the difference between any other pair of adjacent terms. The correct choice is A.|
|2. The sum of the terms in an infinite geometric series can be either finite or
infinite. As an example of an infinite geometric series whose sum is finite (and which, by
definition, converges), consider
A = 1 + 1/2 + 1/4 + 1/8 + 1/16 +
1/32 + ...
As an example of an infinite geometric series whose sum is infinite (and which, by definition, diverges), consider
B = 1 + 2 + 4 + 8 + 16 + 32 + ...
The correct choice is C.
|3. As we increase the value of a positive integer n without limit, the value of n! increases without limit (but faster). Therefore, as n gets larger endlessly, the value of 1/n! approaches 0, just as the value of 1/n does (but faster). The correct choice is D.|
|4. The sequence G given in this example is more interesting than the sequence F
described in Question 4. Remember that 0! = 1 by definition. When we simplify the terms in
G individually, we get the following values:
0!/1! = 1/1
1!/2! = 1 / (1 x 2)
2!/3! = (1 x 2) / (1 x 2 x 3)
3!/4! = (1 x 2 x 3) / (1 x 2 x 3 x 4)
4!/5! = (1 x 2 x 3 x 4) / (1 x 2 x 3 x 4 x 5)
In each element of the series, all of the terms in the numerator cancel out leaving 1, and all of the terms in the denominators cancel out leaving n. In general, we can express the nth term as
(n - 1)! / n! = 1/n
Therefore, we can rewrite the sequence as
G = 1, 1/2, 1/3, 1/4, 1/5, 1/6, ..., 1/n, ...
As n grows without limit, the value of n approaches 0, so the answer is D.
|5. When an infinite series converges, then by definition we can always identify a single, clearly defined limit value. The answer is A.|
|6. At first, you might dread working out this problem, fearing that you'll have to
drag yourself through a lot of tedious arithmetic. However, you can simply look at the
formula for the series expansion of the sine function and see what happens to the term
values as you go out. Once again, that formula is
sin θ = θ - θ3/3! + θ5/5! - θ7/7! + θ9/9! - ...
for any angle θ. In this case, you have θ = π/4 rad, which equals a positive number less than 1. (If you work it out to six decimal places, you'll get π/4 = 0.795398.) As you raise any positive number less than 1 to ever-increasing integer powers, the value approaches 0. Therefore, the numerators in the series expansion for sin π/4 approach 0 as you keep generating new terms. The denominators, on the other hand, skyrocket! You can therefore conclude that the term values approach 0 as you move out along the series. The correct choice is B.
|7. This situation is almost like the one described in Question 6, except that θ = 1 rad. The numerators in the sine function's series expansion therefore remain at 1 in every case. (As you raise 1 to ever-increasing integer powers, the value always equals 1.) The denominators, however, increase without bound, just as they did in the previous case. It follows that the term values once again approach 0 as you move out along the series. The answer is B.|
|8. For any given angle θ, the series expansion for sin θ converges if and only if we can define sin θ. No matter what value we "input" to the sine function, we always get a defined "output" value. Any real number whatsoever will work! Therefore, the series expansion for sin θ converges in all three cases A, B, and C. The correct choice is D, "All of the above."|
|9. For any given angle θ, the series expansion for cos θ converges if and only if we can define cos θ. Just as is the case with the sine function, it doesn't matter what value we "input" to the cosine function; we always get a defined "output" value. The series expansion for cos θ converges in all three cases A, B, and C, so once again the answer is D, "All of the above."|
|10. For any given angle θ, the series expansion for tan θ converges if and only if we can define tan θ. In the range of "input" values 0 ≤ θ ≤ π/2, we get a defined "output" value for anything except π/2. In other words, we can "input" anything in the range 0 ≤ θ < π/2 and expect the series expansion to converge. The correct choice is B.|