If y = 6Σk=1 k cos-1 { 3/5 coskx - 4/5 sin kx }.

Question:

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

then $\frac{\mathrm{dy}}{\mathrm{dx}}$ at $\mathrm{x}=0$ is________.

Solution:

Put $\cos \alpha=\frac{3}{5}, \sin \alpha=\frac{4}{5} \quad 0<\alpha<\frac{\pi}{2}$

Now $\frac{3}{5} \cos \mathrm{kx}-\frac{4}{5} \sin \mathrm{kx}$

$=\cos \alpha \cdot \cos \mathrm{kx}-\sin \alpha \cdot \sin \mathrm{kx}$

$=\cos (\alpha+\mathrm{kx})$

As we have to find derivate at $x=0$

We have $\cos ^{-1}(\cos (\alpha+k x))$

$=(\alpha+\mathrm{kx})$

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

$\left.\Rightarrow \frac{\mathrm{dy}}{\mathrm{dx}}\right|_{a \mathrm{x}=0}=\sum_{k=\mathrm{x}}^{6} \mathrm{k}=\frac{6 \times 7 \times 13}{6}=91$

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