If `y=sum_(k=1)^6 K cos^(-1)(3/5coskx-4/5sinkx)` then `(dy)/(dx)=`

Options

A. 92
B. 91
C. 90
D. 89

Answer

To solve the problem, we need to differentiate the given function \( y \) with respect to \( x \). The function is defined as:

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

### Step 1: Simplifying the Argument of the Inverse Cosine
We can rewrite the argument of the inverse cosine function using the cosine addition formula. We know that:

\[ \cos(\alpha) \cos(\beta) - \sin(\alpha) \sin(\beta) = \cos(\alpha + \beta) \]

Let \( \alpha = \cos^{-1} \left( \frac{3}{5} \right) \) and \( \beta = kx \). Thus, we can express the argument as:

\[ \frac{3}{5} \cos(kx) - \frac{4}{5} \sin(kx) = \cos\left(kx + \alpha\right) \]

Now, substituting this back into the equation for \( y \):

\[ y = \sum_{k=1}^{6} k \cos^{-1} \left( \cos\left(kx + \alpha\right) \right) \]

### Step 2: Using the Property of Inverse Cosine
Since \( \cos^{-1}(\cos(\theta)) = \theta \) for \( \theta \) in the range of \( \cos^{-1} \), we can simplify further:

\[ y = \sum_{k=1}^{6} k \left(kx + \alpha\right) \]

### Step 3: Expanding the Summation
Now we can expand the summation:

\[ y = \sum_{k=1}^{6} k^2 x + \sum_{k=1}^{6} k \alpha \]

### Step 4: Calculating the Summations
1. The first summation \( \sum_{k=1}^{6} k^2 \) can be calculated using the formula:

\[ \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} \]

For \( n = 6 \):

\[ \sum_{k=1}^{6} k^2 = \frac{6(6+1)(2 \cdot 6 + 1)}{6} = \frac{6 \cdot 7 \cdot 13}{6} = 91 \]

2. The second summation \( \sum_{k=1}^{6} k \) is:

\[ \sum_{k=1}^{6} k = \frac{6(6+1)}{2} = 21 \]

Thus, we can write:

\[ y = 91x + 21\alpha \]

### Step 5: Differentiating with Respect to \( x \)
Now, we differentiate \( y \) with respect to \( x \):

\[ \frac{dy}{dx} = 91 \]

### Final Answer
Thus, the final answer for \( \frac{dy}{dx} \) is:

\[ \frac{dy}{dx} = 91 \]

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