Range And Rearrange

Using the Trigonometric-R method twice, $7=2\sin a\sin b +3 \cos b + 6 \cos a \sin b$ $=(2\sin a + 6\cos a)\sin b + 3\cos b$ $\leq \sqrt{(2\sin a + 6\cos a)^2+3^2}$ $\leq \sqrt{(\sqrt{2^2+6^2})^2+3^2}$ $=\sqrt{49}$ $=7$

Hence the LHS has to achieve its maximum when $a+\tan^{-1}{\frac62}=\frac{\pi}{2}$ $\Rightarrow \tan a = \frac13$ $b+\tan^{-1}{\frac{3}{\sqrt{2^2+6^2}}}=\frac{\pi}{2}$ $\Rightarrow \tan b = \frac{2\sqrt{10}}{3}$ $\tan^2 a + 2\tan^2 b = \left(\frac13\right)^2+2\left(\frac{2\sqrt{10}}{3}\right)^2$ $=\boxed9$

My lines of thought for Solution 1

I know $2^2 + 3^2 + 6^2 = 7^2$ , and $(\sin a \sin b, \cos a \sin b, \cos b)$ is the stereographic projection of the sphere, so the above equation looks tight. We should be able to bound it in some way.

We know the identity $\sin^2 \theta + \cos^2 \theta = 1$ . Can we use this to put a bound on $x \sin \theta + y \cos \theta$ ? Of course; one of the simplest ways is to use calculus. Assume $x,y > 0$ to make things simpler. Then clearly the upper bound is obtained when $\sin \theta, \cos \theta > 0$ .

Substitute $\cos \theta = \sqrt{1 - \sin^2 \theta}$ , so we want to maximize $x \sin \theta + y \sqrt{1 - \sin^2 \theta}$ . The maximum is achieved when the derivative with respect to $\sin \theta$ is 0. This derivative is $x - \frac{y \sin \theta}{\sqrt{1 - \sin^2 \theta}}$ , so we have $y \sin \theta = x \sqrt{1 - \sin^2 \theta} = x \cos \theta$ . Thus the maximum is achieved when $\sin \theta : \cos \theta = x : y$ , or when $\sin \theta = \frac{x}{\sqrt{x^2+y^2}}, \cos \theta = \frac{y}{\sqrt{x^2+y^2}}$ . This gives a maximum of $\sqrt{x^2+y^2}$ .

We can also show similarly that the minimum is $-\sqrt{x^2+y^2}$ . Additionally, this can be generalized to any $x,y$ , including negative reals and zero.

Can we use this for the problem? Of course. Let $x = 2 \sin a + 6 \cos a$ , then $x \in [-\sqrt{40}, \sqrt{40}]$ . In addition, given such $x$ , we have $x \sin b + 3 \cos b \le \sqrt{x^2+9}$ . Since $x^2 \le 40$ , we have $\sqrt{x^2+9} \le 7$ . So we do achieve the maximum, which means $x^2 = 40$ . We may take $x = \sqrt{40}$ (the positive one), which leads to $\sin a : \cos a = 2 : 6$ and $\sin b : \cos b = \sqrt{40} : 3$ , leading to $\tan a = \frac{1}{3}, \tan b = \frac{\sqrt{40}}{3}$ , and so $\tan^2 a + 2 \tan^2 b = \frac{1}{9} + 2 \cdot \frac{40}{9} = \boxed{9}$ . The other choice $x = -\sqrt{40}$ leads to the same answer.

How I write Solution 1

Lemma 1 : If $x,y$ are real numbers so that not both are zero, then the maximum of $x \sin \theta + y \cos \theta$ is $\sqrt{x^2+y^2}$ , achieved when $\sin \theta : \cos \theta = x : y$ . (And the minimum is $-\sqrt{x^2+y^2}$ , achieved when $\sin \theta : \cos \theta = -x : y$ .

By Pythagorean theorem, $\cos \theta = \sqrt{1 - \sin^2 \theta}$ . Thus we're looking for the maximum of $x \sin \theta + y \sqrt{1 - \sin^2 \theta}$ . Substitute $s = \sin \theta$ ; we have $s \in [-1, 1]$ . Since $[-1, 1]$ is closed, the function $s \mapsto x s + y \sqrt{1 - s^2}$ does have a maximum (by Extreme Value Theorem). Since it's differentiable everywhere in $(-1, 1)$ , the maximum is achieved either on an endpoint or some point where the derivative is 0.

The derivative can be computed by standard techniques, and is equal to $x - y \frac{s}{\sqrt{1-s^2}}$ . This is zero when $\frac{s}{\sqrt{1-s^2}} = \frac{x}{y}$ . Since $s = \sin \theta, \sqrt{1-s^2} = \cos \theta$ , this is zero when $\sin \theta : \cos \theta = x : y$ . Thus $\sin \theta = \frac{x}{k}, \cos \theta = \frac{y}{k}$ for some $k$ . Since $\sin^2 \theta + \cos^2 \theta = 1$ , we have $\frac{x^2}{k^2} + \frac{y^2}{k^2} = 1$ , or $k = \sqrt{x^2+y^2}$ .

At $s = \frac{x}{\sqrt{x^2+y^2}}$ , the function takes value $\sqrt{x^2+y^2}$ . At the endpoint $s = 1$ , the function takes value $x$ , and at $s = -1$ , it takes value $-x$ . Since $y^2 \ge 0$ , we have $\sqrt{x^2+y^2} \ge \sqrt{x^2} = |x| \ge \pm x$ . Thus the maximum is $\sqrt{x^2+y^2}$ , reached when $\sin \theta : \cos \theta = x : y$ , proving the claim.

The minimum is proven in pretty much the same way.

We will now return to the problem. Let $x = 2 \sin a + 6 \cos a$ . Thus $2 \sin a \sin b + 3 \cos b + 6 \cos a \sin b = x \sin b + 3 \cos b$ . By Lemma 1 with $x \gets 2, y \gets 6, \theta \gets a$ , we have $2 \sin a + 6 \cos a = x \in [-\sqrt{40}, \sqrt{40}]$ and so $x^2 \le 40$ . By Lemma 1 as well with $x \gets x, y \gets 3, \theta \gets b$ , we have $x \sin b + 3 \cos b \le \sqrt{x^2+9}$ . But we have $x \sin b + 3 \cos b = 7$ , so $7 \le \sqrt{x^2+9}$ , or $x^2 \ge 40$ . Thus we have $x = \pm \sqrt{40}$ . We'll take the positive root; it turns out that either choice is fine. Since $x = \sqrt{40}$ , both times Lemma 1 was invoked, they achieved the maximum. Thus $\sin a : \cos a = 2 : 6$ and $\sin b : \cos b = \sqrt{40} : 3$ . Thus $\tan a = \frac{2}{6} = \frac{1}{3}$ and $\tan b = \frac{\sqrt{40}}{3}$ , giving $\tan^2 a + 2 \tan^2 b = \frac{1}{9} + 2 \cdot \frac{40}{9} = \boxed{9}$ .

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My lines of thought for Solution 2

The above feels like a lot of work for the lemma. Can we simplify it somehow? Something involving squaring the individual terms and multiplying corresponding terms... Oho, vectors. (Norm of vector is squaring individual terms, and inner product is multiplying corresponding terms.) Let's do $u = (2, 6, 3), v = (\sin a \sin b, \cos a \sin b, \cos b)$ so we have $u \cdot v = 7$ . Also, since we're working with norms, we have $|u \cdot v| = |u| |v| |\cos \theta|$ , no? Then $|u| = 7, |v| = 1$ , so... $|\cos \theta| = 1$ ? How convenient!

How I write Solution 2

Let $u = (2, 6, 3), v = (\sin a \sin b, \cos a \sin b, \cos b)$ . Then the left hand side is $u \cdot v)$ . The norm is $|u \cdot v| = |u| |v| |\cos \theta|$ where $\theta$ is the angle between the two vectors. We have $|u| = \sqrt{2^2 + 6^2 + 3^2} = 7$ , $|v| = \sqrt{\sin^2 a \sin^2 b + \cos^2 a \sin^2 b + \cos^2 b} = 1$ , and $|u \cdot v| = 7$ from the problem. Thus $|\cos \theta| = 1$ ; the two vectors have the same direction. Thus $\sin a \sin b = 2k, \cos a \sin b = 6k, \cos b = 3k$ for some $k$ . (In fact, since $|v| = 1, |u| = 7$ , we have $|k| = \frac{|v|}{|u|} = \frac{1}{7}$ . This is only important to show that there is indeed a solution for $a,b$ .)

Thus $\tan a = \frac{\sin a \sin b}{\cos a \sin b} = \frac{1}{3}$ . Also, $\sin b = \sqrt{(\sin a \sin b)^2 + (\cos a \sin b)^2} = k\sqrt{40}$ , so $\tan b = \frac{\sin b}{\cos b} = \frac{\sqrt{40}}{3}$ . Thus $\tan^2 a + 2 \tan^2 b = \frac{1}{9} + 2 \cdot \frac{40}{9} = \boxed{9}$ .