If y = sum_{k=1}^{6} k cos^{-1} left{ frac{3} | vedclass

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(B) Let $\cos \alpha = \frac{3}{5}$ and $\sin \alpha = \frac{4}{5}$,where $0 < \alpha < \frac{\pi}{2}$.The expression inside the $\cos^{-1}$ function

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$91$

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(B) Let $\cos \alpha = \frac{3}{5}$ and $\sin \alpha = \frac{4}{5}$,where $0 The expression inside the $\cos^{-1}$ function is $\frac{3}{5} \cos kx - \frac{4}{5} \sin kx$.Using the trigonometric identity $\cos(A+B) = \cos A \cos B - \sin A \sin B$,we get $\cos \alpha \cos kx - \sin \alpha \sin kx = \cos(\alpha + kx)$.Thus,$y = \sum_{k=1}^{6} k \cos^{-1}(\cos(\alpha + kx))$.For values of $x$ near $0$,$\cos^{-1}(\cos(\alpha + kx)) = \alpha + kx$.Therefore,$y = \sum_{k=1}^{6} k(\alpha + kx) = \sum_{k=1}^{6} (k\alpha + k^2 x)$.Differentiating with respect to $x$,we get $\frac{dy}{dx} = \sum_{k=1}^{6} k^2$.At $x = 0$,$\frac{dy}{dx} = \sum_{k=1}^{6} k^2 = 1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2$.Using the sum of squares formula $\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6}$,for $n=6$:$\frac{6(7)(13)}{6} = 91$.

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

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