Tantalizing Trigonometric Equations

Write the first equation as $a = 100 \cos 2u - 25 = f(u),$ and the second equation as $a = -50\sqrt{3} \left( -\frac{1}{2} \sin v + \frac{\sqrt{3}}{2} \cos v \right) = -50 \sqrt{3} \cos (v + \pi/6) = g(v).$ Since $\cos 2x$ ranges from $1$ to $-1$ for $0 \le x \le \pi/2$ , we see that the smallest positive number that satisfies the first condition is given by $u = f^{-1}(a)$ where $f^{-1}(a)$ is chosen such that $0 < u \le \pi/2$ when $-125 \le a < 75$ , and $u = f^{-1}(75) = \pi$ when $a = 75$ . (We cannot choose $u = f^{-1}(75) = 0$ .)

Next, observe that $-50\sqrt{3} \le a < -75$ for $-\pi/3 < x < 0$ , and on this interval, $g(x)$ is symmetric about the line $x = -\pi/6$ ; i.e., $g(-\pi/6+x) = g(-\pi/6-x)$ . So the largest negative number that satisfies the second condition for $a < -75$ must satisfy $-\pi/6 \le v < 0$ . That is, $v = g^{-1}(a)$ is chosen such that $-\pi/6 \le v < 0$ if $-50\sqrt{3} \le a < -75$ , and $-7\pi/6 \le v \le -\pi/3$ if $-75 \le a \le 50\sqrt{3}$ .

The common interval of $a$ for which both inverses are defined is $-50\sqrt{3} \le a \le 75$ . Since $f^{-1}(-75) = \frac{1}{2} \cos^{-1} \frac{-1}{2} = \frac{\pi}{3}$ and $f^{-1}(a)$ is a decreasing one-to-one function on $(-50\sqrt{3}, 75)$ , $u = f^{-1}(a) < \pi/3 < -v$ for $-75 < a < 75$ . Therefore, there are no solutions on this interval for $a$ .

For $a = 75$ , $u + v = \pi + (-\pi) = 0$ , so this is one solution.

For $a = -75$ , $u + v = \pi/3 + (-\pi/3) = 0$ , so this is a second solution.

Hence the remaining case to investigate is the interval $-50\sqrt{3} \le a < -75$ , or equivalently, $-86 \le a < -75$ . In this case, $-\pi/6 < v < 0$ , whereas $u > \pi/3$ , so $u + v > 0$ . There are $11$ integer values of $a$ in this interval, each of which admits a solution. Therefore, the total number of integers $a$ for which $u + v \ge 0$ is $13$ .

We realise that in this question, there is both $\sin$ and $\cos$ so intuitively, we might want to convert all of these such that they are all of the same trigonometric function. Now, remark that $\cos^2 - \sin^2$ brings to mind the double-angle formula for $\cos$ . So we consider to change both of the equations to only involving $\cos$ .

So, keeping the above observations in mind, we can write:

• $a = 100 (\cos^2 u - \sin^2 u) - 25 = 100 \cos 2u - 25$

• $a = - 50 \sqrt{3} ( \frac{\sqrt{3}}{2} \cos v -\frac{1}{2} \sin v) = -50 \sqrt{3} \cos(v + \frac{\pi}{6})$

Now, the second "factorisation" might look pretty convoluted, so here's my motivation. We start off by realising that we need $\cos(v ± w)$ for some $w$ . Recall that $\cos(v±w) = \cos v \cos w ∓\sin v \sin w$ , so it is intuitive to consider the ratio $\frac{ 25 \sqrt{3}}{75}$ . In fact $\frac{25}{75} = \frac{1}{3}$ quickly reminded me of $\frac{1}{2}, \frac{\sqrt{3}}{2}$ and I was sort of done. We kept the equations in radians because degree signs are quite troublesome and hard to keep track of.

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By now, with the restated equations, we realise that we need to deal with "overlap" in the ranges of the two equations, or should I say functions . So, we define the functions $p(x)$ respectively $q(x)$ as the first and second equations. In other words, $p(u) = 100 \cos 2u - 25$ and similar for $q(v)$ .

Firstly, remark that $\cos 2u$ takes on the value $( -1,1)$ when $0 < u ≤ \frac{pi}{2}$ . So, talking about range, $-125 ≤ a < 75 ⇔ 0 < u ≤ \frac{\pi}{2}$ . Now, by the question restraint on smallest positive real number for $u = p^{-1}(a)$ , we see that $u = p^{-1}(75) = \pi$ . (or else it have to be $0$ which violates the restraint)

With a similar strategy in mind, we deal with $q(x)$ . However, this is made simpler if we exploit the symmetry behind $\cos(v+ \frac{\pi}{6})$ , noticing that when $k = - \frac{\pi}{6}$ , $q(k)$ is symmetric about this line. This gives us an incentive to conclude that $-50 \sqrt{3} ≤ a < - 75 ⇔ -\frac{\pi}{6} ≤ v < 0$ and $-75 ≤ a ≤ 5- \sqrt{3} ⇔ - \frac{7 \pi}{6} ≤ v ≤ -\frac{\pi}{3}$ .

So really the main motivation here is to bound $a$ such that the $p^{-1}(a), q^{-1}(a)$ is defined.

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So now, the common interval for both of the inverses is $-50 \sqrt{3} ≤ a ≤ 75$ . Again notice that in this interval, $p^{-1}(x)$ is basically a bijective and monotonically decreasing function. But we notice that however, if $a \in (-50\sqrt{3}, 50 \sqrt{3})$ , then $u = p^{-1}(a) < \frac{\pi}{3} < -v$ so there are no solutions in this particular interval.

We will consider the following remaining cases:

• $a = -75$ . Then $u+v = \frac{\pi}{3} + - \frac{\pi}{3} = 0$ which satisfies the conditions.

• $a = 75$ . Then $u+v = \pi + - \pi = 0$ which satisfies the conditions.

• $-50 \sqrt{3} < a ≤ -75$ . So, in this case $-\frac{\pi}{6} < v < 0$ while $u > \frac{\pi}{3}$ . Whence clearly $u+v > 0$ . For simplicity sake notice that $\lceil -50 \sqrt{3} \rceil = -86$ , so there are $11$ integer values in this range.

This final step is fairly standard - it's just checking the intervals and trying to limit down the choices, and ensuring that $u+v≥ 0$ .

In total, there are $11 + 1 + 1 = \fbox{13}$ values of $a$ .

The first equation can be re-written as $100 cos 2u = a+25$ , so $-125 \leq a \leq 75$ . Using the R-formula, the second equation can be re-written as $50\sqrt{3} \sin (v - \frac{\pi}{3}) = a$ , so $-50\sqrt{3} \leq a \leq 50\sqrt{3}$ . Since a is an integer, $-86 \leq a \leq 75$ .

Note that when $-75 < a < 75$ , $0 < 2u < \frac{2\pi}{3}$ and $-\frac{4\pi}{3} < v - \frac{\pi}{3} < -\frac{2\pi}{3}$ . Thus $u < \frac{\pi}{3}$ and $v < -\frac{\pi}{3}$ , so u+v is negative in this range. When $-86 \leq a < -75$ , $2u > \frac{2\pi}{3}$ and $-\frac{\pi}{2} < v - \frac{\pi}{3} < -\frac{\pi}{3}$ . This will result in positive value of u+v in this range. When $a=75$ , $u = \pi$ and $v = -\pi$ , and when $a=-75$ , $u = \frac{\pi}{3}$ and $v = -\frac{\pi}{3}$ . For both cases u+v=0, and we have to add these cases.

Therefore, there are 13 integers $a$ which will result in a non-negative u+v.

Since $\cos^2 u - \sin^2 u = \cos 2u$ , the first equation is equivalent to $a = 100 \cos 2u - 25$ . From this, we see that $a$ can take any value in the range $[-125, 75]$ . The second equation is equivalent to $a = 50\sqrt{3}(\frac{1}{2}\sin v - \frac{\sqrt{3}}{2}\cos v) = 50\sqrt{3}(\cos \frac{\pi}{3}\sin v - \sin \frac{\pi}{3}\cos v)$ , which simplifies to $a = 50\sqrt{3} \sin (v - \frac{\pi}{3})$ . From this, we see that $a$ can take any value in the range $[-50\sqrt{3}, 50\sqrt{3}]$ . Thus, from both equations, we can conclude that $a$ can take any value in the range $[-50\sqrt{3}, 75]$ . But $a$ is an integer, so $a$ can take all the integer values in the range $[-86, 75]$ .

If $a$ is in the range $(-75,75)$ , then we have $\cos 2u = \frac{a+25}{100}$ being in the range $(-\frac{1}{2}, 1)$ . Since $u$ is the smallest positive real number satisfying the equation, it is easy to see from the graph of $\cos 2u$ that $2u$ is in the range $(0, \cos^{-1} (-\frac{1}{2})) = (0, \frac{2\pi}{3})$ . Thus, $u$ is in the range $(0, \frac{\pi}{3})$ . At the same time, we have $\sin (v - \frac{\pi}{3}) = \frac{a}{50\sqrt{3}}$ being in the range $(-\frac{\sqrt{3}}{2}, \frac{\sqrt{3}}{2})$ . By drawing the graph $\sin (v - \frac{\pi}{3})$ and noting that $\sin (-\frac{\pi}{3}) = -\frac{\sqrt{3}}{2}$ , we can see that $v - \frac{\pi}{3}$ is in the range $(-\sin^{-1} \frac{\sqrt{3}}{2} - \pi, - \sin^{-1} (-\frac{\sqrt{3}}{2}) - \pi) = (-\frac{4\pi}{3}, -\frac{2\pi}{3})$ . Thus, $v$ is in the range $(-\pi , -\frac{\pi}{3})$ . Since $u < \frac{\pi}{3}$ and $v < -\frac{\pi}{3}$ , then $u+v < 0$ and is clearly not non-negative.

If $a$ = $75$ , then it is easy to see that $u = \pi$ and $v = -\pi$ . This gives $u+v = 0$ , so $a = 75$ satisfies the criteria.

If $a$ = $-75$ , then we have $u = \frac{\pi}{3}$ and $v = -\frac{\pi}{3}$ . This also gives $u+v = 0$ , so $a = -75$ satisfies the criteria.

Now consider $a$ in the range $[-86,-75)$ , or rather, $(-50\sqrt{3}, -75)$ . Then we have $\cos 2u$ being in the range $(\frac{1-2\sqrt{3}}{4}, -\frac{1}{2})$ . So $2u > \cos^{-1} (-\frac{1}{2}) = \frac{2\pi}{3} \Rightarrow u > \frac{\pi}{3}$ . Furthermore, $\sin (v - \frac{\pi}{3})$ is in the range $(-1, -\frac{\sqrt{3}}{2})$ , so $v - \frac{\pi}{3}$ must be in the range $(\sin^{-1} (-1), \sin^{-1} (-\frac{\sqrt{3}}{2})) = (-\frac{\pi}{2}, -\frac{\pi}{3})$ . This means that $v$ is in the range $(-\frac{\pi}{6}, 0)$ . Since $u > \frac{\pi}{3}$ and $v > -\frac{\pi}{6}$ , then $u+v > \frac{\pi}{6} > 0$ , so all $a$ in the range $[-86,-75)$ will result in a positive value for $u+v$ .

Summarising the above, $a$ can only take either the value $75$ or all values in the range $[-86,-75]$ . There are $13$ such integer values, so the answer is $13$ .

We want $u + v$ to be non-negative, which can be represented mathematically by:

$u + v \ge 0$

Since $u$ is positive, $u = |u|$ ,

and since $v$ is negative,

$|v| = - v$

$v = - |v|$

$u + v = |u| - |v|$

Therefore, there required condition is,

$|u| - |v| \ge 0$

$|u| \ge |v|$

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Now we shall try to simply the given equations. The first equation is

$100$ cos $^{2} u - 100$ sin $^{2} u = a + 25$

$100$ $($ cos $^{2} u -$ sin $^{2} u ) = a + 25$

Using the identity: sin $^{2} u +$ cos $^{2} u = 1$ , we get

$100$ $(2$ cos $^{2} u - 1) = a + 25$

On further simplifying, we get

cos $^{2} u = \frac{a + 125}{200}$

Note that the range of cos $^{2} u$ is $[0, 1]$ , so the range of $a$ resulting in a real value of $u$ is $[ -125, 75]$

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Now let us see the second equation.

$25\sqrt{3}$ sin $v - 75$ cos $v = a$

$\frac{50\sqrt{3}}{2}$ sin $v - \frac{150}{2}$ cos $v = a$

$50\sqrt{3} ( \frac{1}{2}$ sin $v - \frac{\sqrt{3}}{2}$ cos $v )= a$

$50\sqrt{3} ($ sin $(\frac{\pi}{6})$ sin $v -$ cos $(\frac{\pi}{6})$ cos $v )= a$

$- 50\sqrt{3} ($ cos $(\frac{\pi}{6})$ cos $v -$ sin $(\frac{\pi}{6})$ sin $v)= a$

$- 50\sqrt{3}$ cos $(\frac{\pi}{6} + v) = a$

cos $(\frac{\pi}{6} + v) = \frac{- a\sqrt{3}}{150}$

Note that the range of cos $(\frac{\pi}{6} + v )$ is $[-1,1]$ , so the range of $a$ resulting in a real value of $v$ is $[ -50\sqrt{3}, 50\sqrt{3}]$ .

As $a$ is the same in both equations, to get the final range of $a$ , we will take the intersection of the above two ranges, to get $[-50\sqrt{3}, 75]$

However, since $a$ is an integer, $a$ lies between $-86$ and $75$ (both inclusive).

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Solving the first equation,

cos $u = \pm \sqrt{\frac{a + 125}{200}}$

Note that the minimum positive value of $u$ will always lie between $(0, \frac{\pi}{2})$ for all values of $a$ except when $a = 75$ .

So, we can make two cases: $a = 75$ and $a < 75$

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Case $I$ : $a = 75$

From the first equation, cos $u = \pm 1$

Therefore, cos $u = 0$ or $\pi$ . Since $u$ is a smallest positive integer, $u \neq 0$ ,

Therefore, $u = \pi$

Now, let's see whether $a= 75$ satisfies the second equation

cos $(\frac{\pi}{6} + v) = \frac{- 75\sqrt{3}}{150}$

cos $(\frac{\pi}{6} + v)= \frac{- \sqrt{3}}{2}$

$\frac{\pi}{6} + v = - \frac{5\pi}{6}$

$v = - \pi$

$|u| = |v|$ , Therefore, $a = 75$ is acceptable.

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Case $II$ : $a < 75$

$|u| < |v|$ for $- 75 < a < 75$

$|u| = |v|$ for $a = - 75$

$|u| > |v|$ for $-86 \le a < - 75$

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So, finally, the satisfying integer values of $a$ are $\{75, -75, -76, -77, -78, -79, -80, -81, -82, -83, -84, -85, -86\}$

The answer is $\boxed{13}$

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I know it's (very) long, but this is how I solved it.

or a=75