# Difference between revisions of "2013 AIME I Problems/Problem 14"

## Problem 14

For $\pi \le \theta < 2\pi$, let

\begin{align*} P &= \frac12\cos\theta - \frac14\sin 2\theta - \frac18\cos 3\theta + \frac{1}{16}\sin 4\theta + \frac{1}{32} \cos 5\theta - \frac{1}{64} \sin 6\theta - \frac{1}{128} \cos 7\theta + \cdots \end{align*}

and

\begin{align*} Q &= 1 - \frac12\sin\theta -\frac14\cos 2\theta + \frac18 \sin 3\theta + \frac{1}{16}\cos 4\theta - \frac{1}{32}\sin 5\theta - \frac{1}{64}\cos 6\theta +\frac{1}{128}\sin 7\theta + \cdots \end{align*}

so that $\frac{P}{Q} = \frac{2\sqrt2}{7}$. Then $\sin\theta = -\frac{m}{n}$ where $m$ and $n$ are relatively prime positive integers. Find $m+n$.

## Solution

### Solution 1

Noticing the $\sin$ and $\cos$ in both $P$ and $Q,$ we think of the angle addition identities: $$\sin(a + b) = \sin a \cos b + \cos a \sin b, \cos(a + b) = \cos a \cos b + \sin a \sin b.$$ With this in mind, we multiply $P$ by $\sin \theta$ and $Q$ by $\cos \theta$ to try and use some angle addition identities. Indeed, we get \begin{align*} P \sin \theta + Q \cos \theta &= \cos \theta + \dfrac{1}{2}(\cos \theta \sin \theta - \sin \theta \cos \theta) - \dfrac{1}{4}(\sin{2 \theta} \sin \theta + \cos{2 \theta} \cos{\theta}) - \cdots \\ &= \cos \theta - \dfrac{1}{4} \cos \theta - \dfrac{1}{8} \sin{2 \theta} - \dfrac{1}{16} \cos{3 \theta} + \cdots \\ &= \cos \theta - \dfrac{1}{2}P \end{align*} after adding term-by-term. Similar term-by-term adding yields $$P \cos \theta + Q \sin \theta = -2(Q - 1).$$ This is a system of equations; rearrange and rewrite to get $$P(1 + 2 \sin \theta) + 2Q \cos \theta = 2 \cos \theta$$ and $$P \cos^2 \theta + Q \cos \theta(2 + \sin \theta) = 2 \cos \theta.$$ Subtract the two and rearrange to get $$\dfrac{P}{Q} = \dfrac{\cos \theta}{2 + \sin \theta} = \dfrac{2 \sqrt{2}}{7}.$$ Then, square both sides and use Pythagorean Identity to get a quadratic in $\sin \theta.$ Factor that quadratic and solve for $\sin \theta = -17/19, 1/3.$ The answer format tells us it's the negative solution, and our desired answer is $17 + 19 = \boxed{036}.$

### Solution 2

Use sum to product formulas to rewrite $P$ and $Q$

$P \sin\theta\ + Q \cos\theta\ = \cos \theta\ - \frac{1}{4}\cos \theta + \frac{1}{8}\sin 2\theta + \frac{1}{16}\cos 3\theta - \frac{1}{32}\sin 4\theta + ...$

Therefore, $P \sin \theta - Q \cos \theta = -2P$

Using $\frac{P}{Q} = \frac{2\sqrt2}{7}, Q = \frac{7}{2\sqrt2} P$

Plug in to the previous equation and cancel out the "P" terms to get: $\sin\theta - \frac{7}{2\sqrt2} \cos\theta = -2$.

Then use the pythagorean identity to solve for $\sin\theta$, $\sin\theta = -\frac{17}{19} \implies \boxed{036}$

### Solution 3

Note that $$e^{i\theta}=\cos(\theta)+i\sin(\theta)$$

Thus, the following identities follow immediately: $$ie^{i\theta}=i(\cos(\theta)+i\sin(\theta))=-\sin(\theta)+i\cos(\theta)$$ $$i^2 e^{i\theta}=-e^{i\theta}=-\cos(\theta)-i\sin(\theta)$$ $$i^3 e^{i\theta}=\sin(\theta)-i\cos(\theta)$$

Consider, now, the sum $Q+iP$. It follows fairly immediately that:

$$Q+iP=1+\left(\frac{i}{2}\right)^1e^{i\theta}+\left(\frac{i}{2}\right)^2e^{2i\theta}+\ldots=\frac{1}{1-\frac{i}{2}e^{i\theta}}=\frac{2}{2-ie^{i\theta}}$$ $$Q+iP=\frac{2}{2-ie^{i\theta}}=\frac{2}{2-(-\sin(\theta)+i\cos(\theta))}=\frac{2}{(2+\sin(\theta))-i\cos(\theta)}$$

This follows straight from the geometric series formula and simple simplification. We can now multiply the denominator by it's complex conjugate to find:

$$Q+iP=\frac{2}{(2+\sin(\theta))-i\cos(\theta)}\left(\frac{(2+\sin(\theta))+i\cos(\theta)}{(2+\sin(\theta))+i\cos(\theta)}\right)$$ $$Q+iP=\frac{2((2+\sin(\theta))+i\cos(\theta))}{(2+\sin(\theta))^2+\cos^2(\theta)}$$

Comparing real and imaginary parts, we find: $$\frac{P}{Q}=\frac{\cos(\theta)}{2+\sin(\theta)}=\frac{2\sqrt{2}}{7}$$

Squaring this equation and letting $\sin^2(\theta)=x$:

$\frac{P^2}{Q^2}=\frac{\cos^2(\theta)}{4+4\sin(\theta)+\sin^2(\theta)}=\frac{1-x^2}{4+4x+x^2}=\frac{8}{49}$

Clearing denominators and solving for $x$ gives sine as $x=-\frac{17}{19}$.

$017+019=\boxed{036}$

### Solution 4

A bit similar to Solution 3. We use $\phi = \theta+90^\circ$ because the progression cycles in $P\in (\sin 0\theta,\cos 1\theta,-\sin 2\theta,-\cos 3\theta\dots)$. So we could rewrite that as $P\in(\sin 0\phi,\sin 1\phi,\sin 2\phi,\sin 3\phi\dots)$.

Similarly, $Q\in (\cos 0\theta,-\sin 1\theta,-\cos 2\theta,\sin 3\theta\dots)\implies Q\in(\cos 0\phi,\cos 1\phi, \cos 2\phi, \cos 3\phi\dots)$.

Setting complex $z=q_1+p_1i$, we get $z=\frac{1}{2}\left(\cos\phi+\sin\phi i\right)$

$(Q,P)=\sum_{n=0}^\infty z^n=\frac{1}{1-z}=\frac{1}{1-\frac{1}{2}\cos\phi-\frac{i}{2}\sin\phi}=\frac{1-0.5\cos\phi+0.5i\sin\phi}{\dots}$.

The important part is the ratio of the imaginary part $i$ to the real part. To cancel out the imaginary part from the denominator, we must add $0.5i\sin\phi$ to the numerator to make the denominator a difference (or rather a sum) of squares. The denominator does not matter. Only the numerator, because we are trying to find $\frac{P}{Q}=\tan\text{arg}(\Sigma)$ a PROPORTION of values. So denominators would cancel out.

$\frac{P}{Q}=\frac{\text{Re}\Sigma}{\text{Im}\Sigma}=\frac{0.5\sin\phi}{1-0.5\cos\phi}=\frac{\sin\phi}{2-\cos\phi}=\frac{2\sqrt{2}}{7}$.

Setting $\sin\theta=y$, we obtain $$\frac{\sqrt{1-y^2}}{2+y}=\frac{2\sqrt{2}}{7}$$ $$7\sqrt{1-y^2}=2\sqrt{2}(2+y)$$ $$49-49y^2=8y^2+32y+32$$ $$57y^2+32y-17=0\rightarrow y=\frac{-32\pm\sqrt{1024+4\cdot 969}}{114}$$ $$y=\frac{-32\pm\sqrt{4900}}{114}=\frac{-16\pm 35}{57}$$.

Since $y<0$ because $\pi<\theta<2\pi$, $y=\sin\theta=-\frac{51}{57}=-\frac{17}{19}$. Adding up, $17+19=\boxed{036}$.

### Solution 5 (lots of room for sillies, I wouldn't recommend it)

We notice $\sin\theta=\frac{-i}{2}(e^{i\theta}-e^{-i\theta})$ and $\cos\theta=\frac{1}{2}(e^{i\theta}+e^{-i\theta})$

With these, we just quickly find the sum of the infinite geometric series' in $P$ and $Q$. $P$ has 2 parts, the $\cos$ and the $\sin$ parts. The $\cos$ part is: $\frac12\cos\theta-\frac18\cos3\theta+\cdots$, which can be turned into: $\frac14(e^{i\theta}(1-\frac{e^{i2\theta}}{4}+\cdots)+e^{-i\theta}(1-\frac{e^{-i2\theta}}{4}+\cdots))$, which is $\frac{1}{4}(\frac{e^{i\theta}}{1+\frac{1}{4}e^{i2\theta}}+\frac{e^{-i\theta}}{1+\frac{1}{4}e^{-i2\theta}})$. This turns into $\frac{5(e^{i\theta}+e^{-i\theta})}{17+4e^{i2\theta}+4e^{-i2\theta}}$.

Following the same process as above, we find that the $\sin$ part of $P$ is $\frac{2i(e^{i2\theta}-e^{-i2\theta})}{17+4e^{i2\theta}+4e^{-i2\theta}}$, the $\cos$ part of $Q$ is $\frac{16+2(e^{i2\theta}+e^{-i2\theta})}{17+4e^{i2\theta}+4e^{-i2\theta}}$, and finally, the $\sin$ part of $Q$ is $\frac{3i(e^{i\theta}-e^{-i\theta})}{17+4e^{i2\theta}+4e^{-i2\theta}}$.

We convert all 4 of these equations into trig, and we end up getting $\frac{2\sqrt{2}}{7}=\frac{10\cos{\theta}-4\sin{2\theta}}{16+4\cos{2\theta}-6\sin{\theta}}$, we divide by $2$ on both numerator and denominator, and we get $\frac{2\sqrt{2}}{7}=\frac{5\cos{\theta}-2\sin{2\theta}}{8+2\cos{2\theta}-3\sin{\theta}}$. We use some trig identities and we get $\frac{\cos{\theta}(5-4\sin{\theta})}{10-4\sin^2{\theta}-3\sin{\theta}}=\frac{2\sqrt2}{7}$. We factor the denominator into $(5-4\sin\theta)(2+\sin\theta)$. We cancel out $5-4\sin\theta$ on both numerator and denominator to get $\frac{\cos\theta}{2+\sin\theta}=\frac{2\sqrt2}{7}$. We set $\sin\theta$ as $x$, and we just solve a quadratic in terms of $x$, $\frac{1-x^2}{x^2+4x+4}=\frac{8}{49}$, cross multiply and simplify, and we get $57x^2+32x-17=0$. We can actually factor this to get $(19x+17)(3x-1)=0$, which yields the 2 solutions $x=-\frac{17}{19}, x=\frac{1}{3}$. Since $\pi\leq\theta<2\pi$, the latter solution is deemed invalid, and we are left with $\sin\theta=-\frac{17}{19}$. Our final answer is $17+19=\boxed{036}$.

~ASAB

omg this was definitely the hardest problem on the AIME I

### Solution 6

Follow solution 3, up to the point of using the geometric series formula $$Q+iP=\frac{1}{1+\frac{\sin(\theta)}{2}-\frac{Qi\cos(\theta)}{2}}$$

Moving everything to the other side, and considering only the imaginary part, we get $$Pi+\frac{Pi}{2}\sin\theta-\frac{Qi}{2}\cos\theta = 0$$

We can then write $P = 2 \sqrt{2} k$, and $Q = 7k$, ($k \neq 0$). Thus, we can substitute and divide out by k. $$2\sqrt{2}+\sqrt{2}\sin\theta-\frac{7}{2}\cos\theta\ =\ 0$$ $$2\sqrt{2}+\sqrt{2}\sin\theta-\frac{7}{2}\sqrt{1-\sin^{2}\theta}=\ 0$$ $$2\sqrt{2}+\sqrt{2}\sin\theta\ =\frac{7}{2}\left(\sqrt{1-\sin^{2}\theta}\right)$$ $$8+8\sin\theta+2\sin^{2}\theta=\frac{49}{4}-\frac{49}{7}\sin^{2}\theta$$ $$\frac{57}{4}\sin^{2}\theta+8\sin\theta-\frac{17}{4} = 0$$ $$57\sin^{2}\theta+32\sin\theta-17 = 0$$ $$\left(3\sin\theta-1\right)\left(19\sin\theta+17\right) = 0$$

Since $\pi \le \theta < 2\pi$, we get $\sin \theta < 0$, and thus, $\sin\theta = \frac{-19}{17} \implies \boxed{036}$

-Alexlikemath