Animated Solution for Mathematics - Trigonometry: Let $\cos (\alpha +\beta )=-\frac{1}{10}$ and $\sin (\alpha -\beta )=\frac{3}{8}$, where $0<\alpha <\frac{\pi }{3}$ and $0<\beta <\frac{\pi }{4}$. If $\tan 2\alpha =\frac{3(1-r\sqrt{5})}{\sqrt{11}(s+\sqrt{5})},r,s\in \mathbb{N}$, then $r+s$ is equal to ......... .

Unlocking the Angle

Welcome, future engineer. Today, we are not just solving a trigonometry problem; we are embarking on a journey of geometric intuition.

When you look at $\tan 2\alpha$, your first instinct might be to reach for the double-angle formula:

But wait—do we know $\tan \alpha$? No. We are given information about $(\alpha +\beta )$ and $(\alpha -\beta )$.

This is the core of the problem: we must decompose $2\alpha$ into these known building blocks. By writing $2\alpha =(\alpha +\beta )+(\alpha -\beta )$, we transform a seemingly impossible task into a beautiful application of the tangent addition formula:

Navigating the Unknown

Before we touch any algebra, we must be detectives. The signs of our trigonometric functions are dictated by the quadrants in which our angles reside.

We are given $0<\alpha <\pi /3$ and $0<\beta <\pi /4$. Adding these, we find $0<\alpha +\beta <7\pi /12$.

The problem tells us $\cos (\alpha +\beta )=-1/10$. Since the cosine is negative, our angle $(\alpha +\beta )$ cannot be in the first quadrant; it must be in the second.

Similarly, for $(\alpha -\beta )$, the range is $-\pi /4<\alpha -\beta <\pi /3$. Given $\sin (\alpha -\beta )=3/8$, which is positive, the angle must be in the first quadrant. This quadrant analysis is the difference between a correct answer and a sign error that ruins everything.

Building the Foundation

Now, let us construct our values. For $(\alpha +\beta )$, we have a cosine of $-1/10$.

Imagine a right triangle where the adjacent side is $1$ and the hypotenuse is $10$. By the Pythagorean theorem, the opposite side is $\sqrt{10^2-1^2}=\sqrt{99}=3\sqrt{11}$.

Since we are in the second quadrant, the tangent is negative:

For $(\alpha -\beta )$, we have a sine of $3/8$. The opposite side is $3$, the hypotenuse is $8$, and the adjacent side is $\sqrt{8^2-3^2}=\sqrt{55}$.

In the first quadrant, the tangent is positive:

Bringing It All Together

Now, we substitute these values into our addition formula:

This looks intimidating, but let us break it down. The numerator is $-3\sqrt{11}+3/\sqrt{55}$. Since $\sqrt{55}=\sqrt{5}\sqrt{11}$, we can write this as:

The denominator becomes:

When we divide the numerator by the denominator, the $\sqrt{5}$ terms cancel out beautifully, leaving us with:

Comparing this to the form $\frac{3(1-r\sqrt{5})}{\sqrt{11}(s+\sqrt{5})}$, we see clearly that $r=11$ and $s=9$. The final step is simple arithmetic:

You have conquered the problem!