Animated Solution for Mathematics - Trigonometry: If $\cot x=\frac{5}{12}$ for some $x\in (\pi ,\frac{3\pi }{2})$, then $\sin 7x\left(\cos \frac{13x}{2}+\sin \frac{13x}{2}\right)+\cos 7x\left(\cos \frac{13x}{2}-\sin \frac{13x}{2}\right)$ is equal to

The expression provided is:

Your first instinct might be panic, but your first action should be expansion. When we multiply these terms out, we get four distinct parts:

Now, look closely for the hidden symmetry. We can group these into two pairs:

This is the moment of clarity. The first group is the classic expansion of $\cos (A-B)$, and the second is the expansion of $\sin (A-B)$.

With $A=7x$ and $B=\frac{13x}{2}$, the entire expression collapses into:

A quick subtraction of the angles, $7x-\frac{13x}{2}=\frac{14x-13x}{2}=\frac{x}{2}$, reveals that our monster has shrunk to the elegant:

Now that we have simplified the expression, we must find the values of $\cos \frac{x}{2}$ and $\sin \frac{x}{2}$. We are given $\cot x=\frac{5}{12}$ and $x\in (\pi ,\frac{3\pi }{2})$.

This places $x$ in the 3rd quadrant, where both sine and cosine are negative. Using the definition of cotangent as $\frac{\text{Base}}{\text{Perpendicular}}$, we visualize a right triangle with base 5 and perpendicular 12. By the Pythagorean theorem, the hypotenuse is $\sqrt{5^2+12^2}=13$, thus $\cos x=-\frac{5}{13}$.

We must determine the quadrant of $\frac{x}{2}$. Since $\pi <x<\frac{3\pi }{2}$, dividing by 2 gives:

This is the 2nd quadrant. In the 2nd quadrant, cosine is negative and sine is positive.

We use the half-angle formulas:

Substituting $\cos x=-\frac{5}{13}$, we find:

Taking the square root and applying the negative sign for the 2nd quadrant, we get $\cos \frac{x}{2}=-\frac{2}{\sqrt{13}}$. Similarly:

Taking the positive square root, we get $\sin \frac{x}{2}=\frac{3}{\sqrt{13}}$. Finally, adding these together:

The final answer is $\frac{1}{\sqrt{13}}$.