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An infinite series pickle

One of the most important topics in your IB Math curriculum is sequences and series. You not only study how sequences of numbers can be related, but also how to apply such in real-world applications/patterns. More often than not, you'll see sequences and series as being part of another topic's exercise (in this case, trigonometry).

Question 🤔

[Maximum marks: 10]

Consider the following infinite sequence:

\sin x \qquad \sin (2x) \qquad 4\sin x -4\sin^3 x  \qquad \cdots

i. Find the ratio of the sequence. [2 marks]

ii. Find the values of x for which the infinite series converges to a sum. [3 marks]

b. Find the sum of the series in terms of $\sin x$ and $\cos x$ . [1 mark]

c. Find the exact value(s) of $\cos x$ for which the sum of the series is 2. [4 marks]

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Solution 🤓

a.i. Given that

\sin (2x) = 2 \sin x \cos x

and that

4 \sin x - 4 \sin^3 x = 4\sin x \left( 1 - \sin^2 x \right) = 4 \sin x \cos^2 x

we can evaluate the ratio using

r = \frac{2 \sin x \cos x}{\sin x } = \frac{4 \sin x \cos^2 x}{2 \sin x \cos x} = 2 \cos x

a.ii. For an infinite series to converge to a sum, its ratio must be between -1 and 1 (not including the aforementioned values). That is,

-1 < 2 \cos x < 1

Dividing all terms by 2,

- \frac{1}{2} < \cos x < \frac{1}{2}

Given that no domain is specified, we must find a general solution for the inequalities below. That region can be found by starting with an odd number of $\frac{\pi}{2}$ turns in the unit circle followed by an addition or subtraction of any arc between 0 and $\frac{\pi}{6}$ .

Each of these $\frac{\pi}{2}$ turns can be represented as

x = (2k+1) \cdot \frac{\pi}{2} \qquad k \in \mathbb{Z}

and each $\frac{\pi}{6}$ decrease or increase can be represented respectively as

x = (2k+1) \cdot \frac{\pi}{2} - \frac{\pi}{6} \qquad k \in \mathbb{Z}

x = (2k+1) \cdot \frac{\pi}{2} + \frac{\pi}{6} \qquad k \in \mathbb{Z}

Representing such solutions graphically would yield the following diagram. Notice that all angles that have a cosine between $- \frac{1}{2}$ and $\frac{1}{2}$ . One of the ways you could write a generalized solution is

\frac{(2k+1) \cdot \pi}{2} - \frac{\pi}{6} < x < \frac{(2k+1) \cdot \pi}{2} + \frac{\pi}{6} \qquad k \in \mathbb{Z}

b. The infinite sum is given by

S_\infty = \frac{u_1}{1-r} = \frac{\sin x}{1 - 2 \cos x}

c. Given that the infinite sum must equal 2,

\frac{\sin x}{1 - 2 \cos x} = 2

\sin x  = 2 - 4 \cos x

Squaring both sides of the equation,

\sin^2 x = 4 - 16\cos x + 16\cos^2 x

Using the trigonometric identity $\sin^2 x + \cos^2 x = 1$

1 - \cos^2 x = 4 - 16 \cos x + 16\cos^2 x

Combining like terms,

17\cos^2 x - 16\cos x + 3 = 0

You may solve this equation performing a variable substitution ( $\cos x$ by $y$ for example) and solving the quadratic equation obtained (the coefficients being 17, -16 and 3 respectively.

\cos x = \frac{- (-16) \pm \sqrt{256-4\cdot 17 \cdot 3}}{2 \cdot 17} = \frac{8 \pm \sqrt{13}}{17}

Given that, as stated previously, the value of $\cos x$ needs to be between $- \frac{1}{2}$ and $\frac{1}{2}$ , the only solution is

\cos x = \frac{8 - \sqrt{13}}{17} \approx 0.258

Using the inverse of cosine,

x \approx 1.31 \ \text{rad}