Matt's Recurrence

Using the relation $\sin(x \pm y)=\sin x \cos y \pm \cos x \sin y$ , one can write:

$\sin((n+2) \theta)=\sin((n+1) \theta) \cos \theta+\cos((n+1)\theta) \sin \theta$

$=2 \sin((n+1) \theta) \cos \theta-\cos \theta \sin((n+1)\theta)+\sin \theta \cos((n+1)\theta)$

$=2 \sin((n+1) \theta) \cos \theta-\sin(n\theta)$

Hence $k_1=2 \cos \theta=2 \sqrt{1-\sin^2 \theta}=2\sqrt{1-\left(\frac{7}{25}\right)^2}=\frac{48}{25}$ and $k_0=-1$ . So, $\frac{p}{q}=\frac{48}{25}-1=\frac{23}{25}$ and $p+q=23+25=48$

As $\sin(\theta) = \frac{7}{25}$ , then $\cos(\theta) = \frac{24}{25}$ . Now, we also have

$(\sin((n+2)\theta) + \sin(n\theta) = 2\sin((n+1)\theta)\cos(\theta)$ $a_{n+2} + a_n = 2 a_{n+1} \cdot \frac{24}{25}$ $a_{n+2} = \frac{48}{25} a_{n+1} - a_n$

Hence, $k_1 = \frac{48}{25}, k_0 = -1$ and $k_1+k_0 = \frac{23}{25}$ from which it follows that $p = 23, q = 25$ and $p+q = 48$ .

Using the factor formula, $sin(n \theta )+sin((n-2) \theta) = 2sin((n-1) \theta )cos(\theta )$ $\Rightarrow sin(n \theta )= (2cos(\theta ))sin((n-1) \theta )+(-1)(sin((n-2) \theta)$ Hence, $k_0 = 2cos(\theta ) = 2\sqrt{1-sin^2(\theta )} = 2 \sqrt{1-(\frac{7}{25})^2} = \frac{48}{25}$ since $\theta$ is acute from range and $k_1 = -1$ . Therefore, $k_0+k_1 = \frac{48}{25}-1=\frac{23}{25}$ so answer is $23+25=48$

Note that if $\sin\theta=\frac{7}{25}$ , then $\cos^2\theta=1-(\frac{7}{25})^2=(\frac{24}{25})^2$ , and since $\pi/2<\theta\le \pi/2$ , we have $\cos\theta\ge 0$ . Hence $\cos\theta=\frac{24}{25}$ .

Using the compound angle formulae for $\sin$ we obtain $\sin(n+2)\theta=\sin(n+1)\theta\cos\theta+\sin\theta\cos(n+1)\theta=\frac{24}{25}a_{n+1}+\sin\theta\cos(n+1)\theta$

Now we use a trick and recognize the last term as part of $\sin((n+1)-1)\theta$ . Hence $\sin\theta\cos(n+1)\theta=(\sin \theta\cos(n+1)\theta-\cos\theta\sin(n+1)\theta)+\cos\theta\sin(n+1)\theta=\frac{24}{25}\sin(n+1)\theta-\sin n\theta$ and $a_{n+2}=\frac{24}{25}a_{n+1}+\frac{24}{25}a_{n+1}-a_n=\frac{48}{25}a_{n+1}-a_n.$ So, $k_1=\frac{48}{25}$ , $k_0=-1$ , $k_1+k_2=\frac{23}{25}$ , and the answer is $48$ .

The recurrence relation $a_{n+2} = k_1a_{n+1} + k_0a_n$ has the general solution $a_n = A_1c_1^n + A_2c_2^n$ , so in this case $a_n = \sin(n\theta) = \frac1{2i}(e^{n\theta i} - e^{-n\theta i})\ \ \therefore\ \ \ c_1 = e^{\theta i},\ \ c_2 = e^{-\theta i}.$

The bases $c_{1,2}$ are the solutions of the characteristic equation $c^2 - k_1c - k_0 = 0,$ , which here is equivalent to $0 = (c-c_1)(c-c_2) \equiv c^2 - k_1c - k_0\ \ \ \therefore\ \ \ k_1 = c_1 + c_2,\ k_2 = -c_1c_2.$ Thus $k_1 = e^{\theta i} + e^{-\theta i} = 2\cos\theta = 2\sqrt{1-(\tfrac7{25})^2} = 2\cdot \frac{24}{25};$ $k_2 = -e^{\theta i}e^{-\theta i} = -e^{\theta i - \theta i} = -1.$ The answer is therefore $k_1 + k_2 = 2\cdot \frac{24}{25} - 1 = \frac{13}{25}\ \ \ \Longrightarrow\ \ \ \boxed{48}.$

This question boils down to a system of equation.

If $\sin \theta = \frac {7}{25}$ , using the Pythagorean theorem, we arrive at $\cos \theta = \frac {24}{25}$ .

Using the double angle identity:

$\sin 2\theta = 2\sin \theta \cos\theta$

$\sin 2\theta= 2 \times \frac {7}{25} \times \frac {24}{25}$

$\sin 2\theta= \frac {336}{625}$

From the equation given:

$\sin 2\theta = k_1 \sin \theta + k_0 \sin 0$

$\frac {336}{625} = k_1 \times \frac {7}{25}$

$k_1 = \frac {48}{25}$

Using the triple angle identity:

$\sin 3 \theta = -4 \sin^3 \theta + 3 \sin \theta$

$\sin 3 \theta = -4 \times (\frac {7}{25})^3 + 3 \times \frac {7}{25}$

$\sin 3 \theta = \frac {11753}{15625}$

$\sin 3 \theta = k_1 \sin 2\theta + k_0 \sin \theta$

$\frac {11753}{15625} = \frac {48}{25} \times \frac {336}{625} + k_0 \times \frac {7}{25}$

$k_0 = -1$

$k_0 +k_1 = \frac {48}{25} - 1$

$k_0 +k_1 = \frac {23}{25}$

$p = 23, q = 25, p+q = 48$

The equation says that $Im(Qz^{n+1})=0$ , where $z=e^{i\theta}$ and $Q = k_1 + k_0z^{-1} - z$ . But that will happen for all $n \in \mathbb{N}$ if and only if $Q=0$ . (If $Q \ne 0$ then $Qz$ and $Qz^2$ cannot both be real, since $z$ is non-real.)

So the condition is equivalent to $z= k_1 + k_0 \overline{z}$ . Since $k_1, k_0$ are real, this gives $k_0=-1$ and the equation becomes $k_1= z+ \overline{z} = 2 cos(\theta) =\frac{48}{25}$ (using Pythagoras). Arithmetic gives $48$ as the answer.

We can easily deduce that $sin \theta$ = 7/25, we therefore get $cos\theta$ = 24/25 Now, putting n = 1 and n = 2 in the recurrence relation we get two equations: $a_{n+2}$ = $a_3$ = $sin 3\theta$ = $k_1$ $sin 2\theta$ + $k_0$ $sin \theta$ … (i) &
$a_{n+2}$ = $a_4$ = $sin 4\theta$ = $k_1$ $sin 3\theta$ + $k_0$ $sin 2\theta$ …(ii)

We know, $sin 3\theta$ = 3 $sin \theta$ – 4 $sin^3 \theta$ and $sin 2\theta$ = 2 $sin \theta$ $sin \theta$ and $sin 4\theta$ = 4 $sin \theta$ $sin \theta$ [1-2 $sin^2\theta$ ]

Substituting values of $sin \theta$ and $cos \theta$ in (i) and (ii), we get,

11753 = 8400 $k_1$ + 4375 $k_0$ => 564144 = 403200 $k_1$ + 210000 $k_0$ [ On multiplying L.H.S & R.H.S by 48 ]…(iii) &

354144=293825 $k_1$ +210000 $k_0$ ...(iv)

Solving (iii) and (iv), we get, $k_1$ = 48/5 and $k_0$ = -1

Therefore, $k_0$ + $k_1$ = 43/5 =p/q Since, gcd (43, 5) = 1 , therefore, p+q= 43+5 = 48.

We have $a_0=0,\ a_1=\sin \theta,\ a_2=\sin 2\theta,\ a_3=\sin 3\theta.$ and because $a_{n+2}=k_1a_{n+1}+k_0a_n,$ so we obtain $\sin 2\theta=k_1\sin \theta$ inferred $k_1=2\cos \theta,$ So, we have $\sin 3\theta=2\cos \theta\sin 2\theta+k_0\sin \theta$ inferred $k_0\sin \theta=\sin 3\theta-2\cos \theta\sin 2\theta=-\sin \theta$ inferred $k_0=-1$ Now, we have $k_0+k_1=2\cos \theta-1=2\sqrt{1-\sin^2 \theta}-1=\frac{23}{25}$ (because $\sin \theta=\frac{7}{25}$ ) Finally, we obtain $p+q=23+25=48$

See, as given in the equation $\theta=sin^{-1} \frac{7}{25}$

and it satisfies, $a {n}=\sin(n \theta)$, $a {n+2}=k {1}a {n+1}+k {0}a {n}$ Now, we find that $a {1}=\frac{7}{25}$ $a {2}=\frac{336}{625}$ $a {3}=\frac{11753}{15625}$ $a {4}=\frac{354144}{390625}$

Forming two equations as, $a {3}=k {1}a {2} + k {0}a {1}$ $a {4}=k {1}a {3} + k {0}a {2}$

solving these using above constraints, we do get, $k {1}=\frac{48}{25}$ and $k {0}=-1$ so, $k {1}+k {0}=\frac{23}{25}$

so, $\boxed {p+q=48}$

We the fact that sin(na)=2sin((n-1) a) cos(a)-sin((n-2)a)

We'll start by calculating some values of $a_n$ .

$a_0 = \sin 0 = 0.$

$a_1 = \sin \theta = \frac{7}{25}.$

$\cos \theta = \sqrt{1-\sin^2 \theta} = \frac{24}{25}.$

$a_2 = \sin 2\theta = 2 \sin \theta \cos \theta = \frac{336}{625}.$

$\cos 2\theta = 1-2\sin^2 \theta = \frac{527}{625}.$

$a_3 = \sin 3\theta = \sin \theta \cos 2\theta + \cos \theta \sin 2\theta = \frac{11753}{15625}.$

Now that we have some values of $a_n$ , we can insert them into the recursive formula to determine $k_1$ and $k_0$ .

First: ${ a }_{ 2 }={ k }_{ 1 }{ a }_{ 1 }+{ k }_{ 0 }{ a }_{ 0 } \Leftrightarrow \frac { 336 }{ 625 } =\frac { 7 }{ 25 } { k }_{ 1 }+0{ k }_{ 0 } \Leftrightarrow { k }_{ 1 }=\frac { 48 }{ 25 }$ .

And finally ${ a }_{ 3 }={ k }_{ 1 }{ a }_{ 2 }+{ k }_{ 0 }{ a }_{ 1 }\Leftrightarrow \frac { 11753 }{ 15625 } =\frac { 48 }{ 25 } \frac { 336 }{ 625 } +{ k }_{ 0 }\frac { 7 }{ 25 } \Leftrightarrow { k }_{ 0 }=-1$ .

So $k_1+k_0=\frac{48}{25}-1=\frac{23}{25}$ . And therefore the answer is $23+25=\fbox{48}$ .