Animated Solution for Mathematics - Differentiation: Let $f(x)=\frac{\sin x+\cos x-\sqrt{2}}{\sin x-\cos x},x\in [0,\pi ]-\{\frac{\pi }{4}\}$, then $f(\frac{7\pi }{12})f^{\prime \prime }(\frac{7\pi }{12})$ is equal to

Imagine you are standing at the base of a massive, jagged mountain. The problem before us, $f(x)=\frac{\sin x+\cos x-\sqrt{2}}{\sin x-\cos x}$, is that mountain.

If you try to climb it by brute-forcing the quotient rule, you will find yourself exhausted, tangled in a web of derivatives, and likely lost in a sea of algebraic errors. But what if there was a hidden path, a secret trail that leads straight to the summit? That is what we are going to find today.

The R-Method

Our first step is to look at the numerator and denominator not as a chaotic mess, but as structured trigonometric entities. We know that any expression of the form $a\sin x+b\cos x$ can be condensed into a single sine wave.

By multiplying and dividing by $\sqrt{a^2+b^2}$, we can rewrite $\sin x+\cos x$ as $\sqrt{2}\sin (x+\frac{\pi }{4})$ and $\sin x-\cos x$ as $\sqrt{2}\sin (x-\frac{\pi }{4})$.

Suddenly, the mountain doesn't look so steep. Our function becomes:

The $\sqrt{2}$ terms cancel out, leaving us with a much cleaner expression. But we can do better.

The Elegant Substitution

To truly simplify, we need to tame the arguments. Let us introduce a new variable, $u=x-\frac{\pi }{4}$. This implies $x+\frac{\pi }{4}=u+\frac{\pi }{2}$.

Substituting this into our function, we get:

Using the allied angle identity $\sin (u+\frac{\pi }{2})=\cos u$, our function transforms into the beautiful, compact form:

The Half-Angle Magic

Now, we reach the most satisfying part of the journey. We use the half-angle identities: $\cos u-1=-2{\sin }^2(\frac{u}{2})$ and $\sin u=2\sin (\frac{u}{2})\cos (\frac{u}{2})$.

When we divide these, the terms cancel with surgical precision:

Substituting $u=x-\frac{\pi }{4}$ back in, we find our final, simplified function: $f(x)=-\tan (\frac{x}{2}-\frac{\pi }{8})$.

The Calculus Sprint

Now that we have a simple tangent function, differentiation is a breeze. The first derivative is $f^{\prime }(x)=-\frac{1}{2}{\sec }^2(\frac{x}{2}-\frac{\pi }{8})$.

Differentiating again, we apply the chain rule to get:

Finally, we evaluate at $x=\frac{7\pi }{12}$. The internal angle becomes $\frac{7\pi }{24}-\frac{3\pi }{24}=\frac{\pi }{6}$.

Evaluating $f(\frac{7\pi }{12})=-\tan (\frac{\pi }{6})=-\frac{1}{\sqrt{3}}$ and $f^{\prime \prime }(\frac{7\pi }{12})=-\frac{1}{2}{\sec }^2(\frac{\pi }{6})\tan (\frac{\pi }{6})=-\frac{1}{2}\cdot (\frac{2}{\sqrt{3}}{)}^2\cdot \frac{1}{\sqrt{3}}=-\frac{2}{3\sqrt{3}}$, we find the product:

And there you have it! By choosing the path of simplification, we conquered the mountain with ease. The final result is $\frac{2}{9}$.