If y = Σ (from k=1 to 6) cos⁻¹ { (3/5) cos kx - (4/5) sin kx }, then dy/dx at x = 0 is

Q: Wait, why do we just cancel $\cos^{-1}$ and $\cos$? Isn't that dangerous?

That is a fantastic question! You are absolutely right to be cautious. Generally, $\cos^{-1}(\cos \theta) = \theta$ only when $\theta$ is in the principal domain $[0, \pi]$. In this problem, we are evaluating the derivative at $x=0$. As $x$ approaches $0$, the angle $kx + \alpha$ approaches $\alpha$. Since $\cos \alpha = 3/5$ and $\sin \alpha = 4/5$, $\alpha$ is clearly an acute angle in the first quadrant, which is well within the $[0, \pi]$ range. So, for the neighborhood of $x=0$, the simplification is perfectly valid.

Q: Why is the derivative of $k\alpha$ zero? Isn't $\alpha$ a variable?

Great catch! In this context, $\alpha$ is a constant defined by the triangle with sides $3, 4, 5$. Specifically, $\alpha = \cos^{-1}(3/5)$. Since it does not depend on $x$, it is treated as a constant. When we differentiate with respect to $x$, any term that is purely a constant (like $k\alpha$) will have a derivative of zero.

Q: How did we know to use the $3-4-5$ triangle?

Pattern recognition is a huge part of JEE Advanced! Whenever you see coefficients like $3/5$ and $4/5$ in a trigonometric expression, your brain should immediately check if their squares sum to $1$ (i.e., $(3/5)^2 + (4/5)^2 = 9/25 + 16/25 = 1$). This is the hallmark of a sine-cosine pair, which allows you to use the compound angle identity. Always look for these 'hidden' triangles!

Question

If $y = \sum_{k=1}^{6} \cos^{-1} \left\{ \frac{3}{5} \cos kx - \frac{4}{5} \sin kx \right\}$, then $\frac{dy}{dx}$ at $x = 0$ is

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing before a mountain of a problem. At first glance, the expression

$$y = \sum_{k=1}^{6} k \cos^{-1} \left( \frac{3}{5} \cos kx - \frac{4}{5} \sin kx \right)$$

looks like a chaotic mess of inverse trigonometric functions. But in the world of JEE Advanced, chaos is often just order in disguise.

### The $3-4-5$ Epiphany

The first thing that should jump out at you is the pair of coefficients: $\frac{3}{5}$ and $\frac{4}{5}$. They are the building blocks of the most famous triangle in geometry—the $3-4-5$ right-angled triangle.

If we define an angle $\alpha$ such that $\cos \alpha = \frac{3}{5}$ and $\sin \alpha = \frac{4}{5}$, we have unlocked the key to the entire problem. This $\alpha$ is an acute angle, sitting comfortably in the first quadrant.

### The Identity Collapse

Now, watch what happens when we substitute these into our expression. The term inside the inverse cosine becomes $\cos \alpha \cos kx - \sin \alpha \sin kx$.

This is the classic expansion of $\cos(A + B)$. The entire expression inside the inverse cosine collapses into $\cos(kx + \alpha)$.

Suddenly, the mountain has become a molehill. Our function is now:

$$y = \sum_{k=1}^{6} k \cos^{-1} (\cos(kx + \alpha))$$

### The Domain Check

This is where many students stumble. We must be careful, as the inverse cosine function is only the identity function in its principal domain, $[0, \pi]$.

Since we are looking for the derivative at $x=0$, we are interested in the behavior of the function near $x=0$. At $x=0$, the angle is $\alpha$. Since $\alpha$ is acute, it is safely within the domain.

Thus, for small $x$, we can simplify the expression to:

$$y = \sum_{k=1}^{6} k(kx + \alpha)$$

### The Final Sprint

Now, the calculus becomes trivial. We distribute the $k$ to get:

$$y = \sum_{k=1}^{6} (k^2 x + k\alpha)$$

When we differentiate with respect to $x$, the $k\alpha$ term vanishes because it is a constant, and the $k^2 x$ term becomes $k^2$. We are left with the derivative:

$$\frac{dy}{dx} = \sum_{k=1}^{6} k^2$$

This is the sum of the first six squares. Using the formula

$$\frac{n(n+1)(2n+1)}{6}$$

with $n=6$, we calculate:

$$\frac{6(7)(13)}{6} = 91$$

And there it is—the elegance of mathematics. A terrifying summation reduced to a single, clean integer. The final answer is **91**.

Animated Solution for Mathematics - Differentiation: If $y = \sum_{k = 1}^{6} cos^{- 1} {\frac{3}{5} cos k x - \frac{4}{5} sin k x}$ , then $\frac{d y}{d x}$ at $x = 0$ is

Enter Numerical Value:

Visualized Solution

Understanding the Function $y$

The $3 - 4 - 5$ Triangle Substitution

Applying the Substitution

The $cos (A + B)$ Identity

Simplifying the Inverse Cosine

Linearizing the Expression for $y$

Differentiating with Respect to $x$

Executing the Derivative

Evaluating the Sum of Squares

Final Calculation and Result

The Sigma Insight: Techniques of Differentiation

Analyzing the Setup

The $3 - 4 - 5$ Epiphany

The Identity Collapse

The Domain Check

The Final Sprint

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Animated Solution for Mathematics - Differentiation: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If y=∑k=16​cos−1{53​coskx−54​sinkx}, then dxdy​ at x=0 is

Enter Numerical Value:

Visualized Solution

Understanding the Function .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y

The .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } 3−4−5 Triangle Substitution

Applying the Substitution

The .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } cos(A+B) Identity

Simplifying the Inverse Cosine

Linearizing the Expression for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } y

Differentiating with Respect to .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x