本帖最后由 hbghlyj 于 2023-3-28 09:51 编辑 C. V. Durell, A. Robson - Advanced Trigonometry-Dover Publications (2003) 第251页
11. Find the sum to infinity of the series,
$$
\cos \alpha \cos \beta-\frac{1}{2} \cos 2 \alpha \cos 2 \beta+\frac{1}{3} \cos 3 \alpha \cos 3 \beta-\ldots,
$$
where neither $\alpha+\beta$ nor $\alpha-\beta$ is an odd multiple of $\pi$.
Solution$\sum_{n=0}^\infty\frac{(-1)^n}{2n}\cos(nx)=\Re\sum_{n=0}^\infty\frac{(-1)^ne^{inx}}{2n}=\Re\frac{ix}2=-\frac{\Im x}2$
$\sum_{n=0}^\infty\frac{(-1)^n}n\cos(n\alpha)\cos(n\beta)=\sum_{n=0}^\infty\frac{(-1)^n}{2n}(\cos(n(\alpha-\beta))-\cos(n(\alpha+\beta)))=$
12. If $x \neq n \pi$, find the sum to infinity of the series,
$$
\cos ^2 x-\frac{1}{2} \sin ^2 2 x+\frac{1}{3} \cos ^2 3 x-\frac{1}{2} \sin ^2 4 x+\ldots
$$
13. If $y=x \sin \alpha+\frac{1}{2} x^2 \sin 2 a+\frac{1}{3} x^3 \sin 3 \alpha+\ldots$, where $|x|<1$, prove that $\sin y=x \sin (y+a)$.
14. If $y=x-t \sin 2 x+\frac{1}{2} t^2 \sin 4 x-\frac{1}{3} t^3 \sin 6 x+\ldots$, whero $t=\tan ^2 \phi<1$, prove that $\tan y=\cos 2 \phi \tan x$.
15. If $|x|<1$, find the sum to infinity of the series,
$$
x \sin \theta+\frac{1}{3} x^3 \sin 3 \theta+\frac{1}{5} x^5 \sin 5 \theta+\ldots
$$
16. If $\frac{\pi}{4}>\alpha>-\frac{\pi}{4}$, prove that
(i) $\tan a \sin \theta-\frac{1}{3} \tan ^3 \alpha \sin 3 \theta+\frac{1}{5} \tan ^5 \alpha \sin 5 \theta-\ldots$
$$
=\frac{1}{4} \log \frac{1+\sin 2 a \sin \theta}{1-\sin 2 \alpha \sin \theta}
$$
(ii) $\tan \alpha \cos \theta-\frac{1}{5} \tan ^3 \alpha \cos 3 \theta+\frac{3}{6} \tan ^5 \alpha \cos 5 \theta-\ldots$ $=\frac{1}{2} \tan ^{-1}(\tan 2 \alpha \cos \theta)$.
17. If $(1+x) \tan \theta=(1-x) \tan \phi$, and if $|x|<1$, expand $\theta$ in ascending powers of $x$.
18. If $\tan \alpha=\cos 2 \omega \tan \lambda$, and $\tan ^2 \omega<1$, expand $\lambda$ in ascending powers of $\tan ^2 \omega$.
19. If $\frac{\pi}{2}<\theta<\frac{3 \pi}{2}$, find the sum to infinity of the series
$$
\cos \theta-\frac{1}{3} \cos 3 \theta+\frac{1}{6} \cos 5 \theta-\ldots
$$
Solution$\cos \theta-\frac{1}{3} \cos 3 \theta+\frac{1}{6} \cos 5 \theta-\ldots=\Re(e^{i\theta}-\frac13e^{3i\theta}+\frac16e^{5i\theta}-\ldots)=\Re(\sin(e^{i\theta}))=\sin(\cos\theta)\cosh(\sin\theta)$
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