Question

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

Hint:

Write the bracket as \(\cos(kx+\alpha)\) with \(\cos\alpha=\frac35,\ \sin\alpha=\frac45\).

Answer 1

Correct Answer: 91

Let \(\cos\alpha=\frac35,\ \sin\alpha=\frac45\) with \(\alpha\in\left(0,\frac{\pi}{2}\right)\).
Then \(\frac35\cos kx-\frac45\sin kx=\cos(kx+\alpha)\).
Near \(x=0\), \(kx+\alpha\) lies in \((0,\pi)\), so \(\cos^{-1}(\cos(kx+\alpha))=kx+\alpha\).
\(y=\sum_{k=1}^{6}k(kx+\alpha)\Rightarrow \frac{dy}{dx}=\sum_{k=1}^{6}k^2\).
\(\sum_{k=1}^{6}k^2=\frac{6\cdot7\cdot13}{6}=91\).
Answer: 91