Animated Solution for Mathematics - Trigonometry: Let $f_k(x)=\frac{1}{k}({\sin }^{k}x+{\cos }^{k}x)$ where $x\in R$ and $k\ge 1$. Then $f_4(x)-f_6(x)$ equals

The beauty of algebraic symmetry in trigonometry is a powerful tool for any future engineer. Today, we demystify a problem that often makes students freeze: the appearance of high powers in trigonometric functions.

We are given the function:

Our objective is to find the value of $f_4(x)-f_6(x)$. When you see ${\sin }^{4}x$ or ${\sin }^{6}x$, your first instinct might be to reach for complex integration or reduction formulas. But stop! In the world of JEE, there is almost always a more elegant path.

The secret here is to stop seeing these as trigonometric beasts and start seeing them as algebraic structures.

The Power of Algebraic Identities

Let us tackle $f_4(x)$ first. We examine the term ${\sin }^{4}x+{\cos }^{4}x$.

If we let $a={\sin }^{2}x$ and $b={\cos }^{2}x$, this expression becomes $a^2+b^2$. We know the classic identity:

Substituting our values back in, we get $({\sin }^{2}x+{\cos }^{2}x{)}^2-2{\sin }^{2}x{\cos }^{2}x$. Because ${\sin }^{2}x+{\cos }^{2}x=1$, this simplifies instantly to $1-2{\sin }^{2}x{\cos }^{2}x$.

Multiplying by the $\frac{1}{4}$ outside, we have:

Extending the Logic to $f_6(x)$

Now, let us apply the same philosophy to $f_6(x)$. We examine ${\sin }^{6}x+{\cos }^{6}x$.

Again, let $a={\sin }^{2}x$ and $b={\cos }^{2}x$. This time, the expression is $a^3+b^3$. We recall the identity:

Substituting our values, we get $({\sin }^{2}x+{\cos }^{2}x{)}^3-3{\sin }^{2}x{\cos }^{2}x({\sin }^{2}x+{\cos }^{2}x)$. Since ${\sin }^{2}x+{\cos }^{2}x=1$, this becomes $1^3-3{\sin }^{2}x{\cos }^{2}x(1)$, which is simply $1-3{\sin }^{2}x{\cos }^{2}x$.

Don't forget the $\frac{1}{6}$ outside, so:

The Grand Cancellation

Now, we bring it all together. We need to calculate $f_4(x)-f_6(x)$. Substituting our simplified forms, we have:

Let us expand this carefully. The first term is $\frac{1}{4}-\frac{2}{4}{\sin }^{2}x{\cos }^{2}x$, which is $\frac{1}{4}-\frac{1}{2}{\sin }^{2}x{\cos }^{2}x$. The second term is $-\frac{1}{6}+\frac{3}{6}{\sin }^{2}x{\cos }^{2}x$, which is $-\frac{1}{6}+\frac{1}{2}{\sin }^{2}x{\cos }^{2}x$.

Look at that! The terms involving ${\sin }^{2}x{\cos }^{2}x$ are $-\frac{1}{2}{\sin }^{2}x{\cos }^{2}x$ and $+\frac{1}{2}{\sin }^{2}x{\cos }^{2}x$. They cancel out perfectly, leaving us with just $\frac{1}{4}-\frac{1}{6}$.

Finding a common denominator of $12$, we get $\frac{3}{12}-\frac{2}{12}=\frac{1}{12}$.

The complexity has vanished, leaving behind a simple, elegant fraction. The final answer is $\frac{1}{12}$.