If $y=\sum_{k=1}^{6} k \cos ^{-1}\left\{\frac{3}{5} \cos k x-\frac{4}{5} \sin k x\right\}$, then $\f

Question

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

Professor PC Thomas Classes · Answer

(91) : We have,\ $y=\sum_{k=1}^{6} k \cos ^{-1}\left\{\frac{3}{5} \cos k x-\frac{4}{5} \sin k xight\}$\ Let $\frac{3}{5}=\cos \alpha$ and $\frac{4}{5}=\sin \alpha$\ Then,\ $y=\sum_{k=1}^{6} k \cos ^{-1}\{\cos \alpha \cos k x-\sin \alpha \sin k x\}$\ $=\sum_{k=1}^{6} k \cos ^{-1} \cos (k x+\alpha)=\sum_{k=1}^{6} k(k x+\alpha)$\ $=\sum_{k=1}^{6}\left(k^{2} x+k \alphaight)$\ $	herefore \quad \frac{d y}{d x}=\sum_{k=1}^{6} k^{2}=1^{2}+2^{2}+\ldots+6^{2}$ $$ =\frac{6(6+1)(2(6)+1)}{6}=7(13)=91 $$