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Question Number 151858 by mathdanisur last updated on 23/Aug/21

Answered by Olaf_Thorendsen last updated on 23/Aug/21

$$f\left(x\right)=\sin x+\frac{\sin 3x}{3}+\frac{\sin 5x}{5}+\sin \frac{7x}{7}$$$$f^{\prime }\left(x\right)=\cos x+\cos 3x+\cos 5x+\cos 7x$$$$f^{\prime }\left(x\right)=\mathrm{Re}(e^{ix}+e^{3ix}+e^{5ix}+e^{7ix})$$$$f^{\prime }\left(x\right)=\mathrm{Re}\left(e^{ix}\frac{1-e^{8ix}}{1-e^{2ix}}\right)$$$$f^{\prime }\left(x\right)=\mathrm{Re}\left(e^{ix}\frac{e^{4ix}(e^{-4ix}-e^{4ix})}{e^{ix}(e^{-ix}-e^{ix})}\right)$$$$f^{\prime }\left(x\right)=\mathrm{Re}\left(e^{4ix}\frac{\sin 4x}{\sin x}\right)$$$$f^{\prime }\left(x\right)=\frac{\cos 4x.\sin 4x}{\sin x}$$$$f^{\prime }\left(x\right)=\frac{\sin 8x}{2\sin x}$$$$f^{\prime }\left(\frac{\pi }{9}\right)=\frac{\sin \frac{8\pi }{9}}{2\sin \frac{\pi }{9}}=\frac{\sin (\pi -\frac{\pi }{9})}{2\sin \frac{\pi }{9}}=\frac{1}{2}$$

Commented by mathdanisur last updated on 23/Aug/21

$$\mathrm{Thank}\mathrm{You}\mathbf{S}\mathrm{er}$$

Commented by Olaf_Thorendsen last updated on 24/Aug/21

$$\mathrm{Let}{u}_n={aq}^{n-1},\underset{k=1}{\sum ^n}{u}_k=a\frac{1-q^n}{1-q}$$$$\mathrm{here}a=e^{ix},q=e^{2ix},n=4$$$$\left(geometricalsequence\right)$$

Commented by mathdanisur last updated on 24/Aug/21

$$\mathrm{Dear}\mathrm{Ser}$$$${\mathrm{f}}^{{}^{\prime }}\left(x\right)=Re(e^{ix}+e^{3ix}+e^{5ix}+e^{7ix})$$$$=Re\left(e^{ix}\frac{1-e^{8ix}}{1-e^{2ix}}\right)$$$$isthereaready-madeformulafor$$$$thistransition?pleasenotehowitis$$

Commented by mathdanisur last updated on 24/Aug/21

$$\mathrm{Thank}\mathrm{you}\mathrm{Ser}$$

Answered by john_santu last updated on 24/Aug/21

$$\mathrm{f}{}^{\prime }\left(\mathrm{x}\right)=\cos \mathrm{x}+\cos 3\mathrm{x}+\cos 5\mathrm{x}+\cos 7\mathrm{x}$$$$\mathrm{f}{}^{\prime }\left(\mathrm{x}\right)=2\cos 4\mathrm{x}\cos 3\mathrm{x}+2\cos 4\mathrm{x}\cos \mathrm{x}$$$$\mathrm{f}{}^{\prime }\left(\mathrm{x}\right)=2\cos 4\mathrm{x}(\cos 3\mathrm{x}+\cos \mathrm{x})$$$$\mathrm{f}{}^{\prime }\left(\mathrm{x}\right)=4\cos \mathrm{x}\cos 2\mathrm{x}\cos 4\mathrm{x}$$$$\mathrm{f}{}^{\prime }\left(\mathrm{x}\right)=\frac{2\sin 2\mathrm{x}\cos 2\mathrm{x}\cos 4\mathrm{x}}{\sin \mathrm{x}}$$$$\mathrm{f}{}^{\prime }\left(\mathrm{x}\right)=\frac{\sin 4\mathrm{x}\cos 4\mathrm{x}}{\sin \mathrm{x}}=\frac{\sin 8\mathrm{x}}{2\sin \mathrm{x}}$$$$\mathrm{f}{}^{\prime }\left(\frac{\pi }{9}\right)=\frac{\sin \left(\frac{8\pi }{9}\right)}{2\sin \left(\frac{\pi }{9}\right)}=\frac{\sin \left(\frac{\pi }{9}\right)}{2\sin \left(\frac{\pi }{9}\right)}=\frac{1}{2}$$

Commented by mathdanisur last updated on 24/Aug/21

$$\mathrm{Thank}\mathrm{you}\mathbf{S}\mathrm{er}$$

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