2018 DSE | Mathematics (Compulsory) 5** | Mathematics (Algebra and Calculus) 5** | 專教數學CORE及M2 | 考試專家 | 技巧取勝
2019年6月1日星期六
Everyday Math 每日數學 之 Mathematical Induction + Trigonometry + Differentiation #20190601 (+Solution)
Solution:
(a)
For any non-negative integer $ n $ , denoted by $ P(n) $ the proposition that $ \cos x \cos (2x) \cos(4x) ... \cos(2^n x)=\frac{\sin(2^{n+1}x)}{2^{n+1}\sin x} $ .
Note that $\cos x=\frac{2\sin x \cos x}{2\sin x}=\frac{\sin(2x)}{2\sin x}$ .
$ P(0) $ is true.
Let $k$ be a non-negative integer. Assume $ P(k) $ is true.
Therefore, $ \cos x \cos (2x) \cos(4x) ... \cos(2^k x)=\frac{\sin(2^{k+1}x)}{2^{k+1}\sin x} $ .
Note that $\cos x \cos (2x) \cos(4x) ... \cos(2^k x)\cos(2^{k+1}x)=\frac{\sin(2^{k+1}x)\cos(2^{k+1}x)}{2^{k+1}\sin x}=\frac{\sin(2^{k+2}x)}{2^{k+2}\sin x}$
Therefore, $ P(k+1) $ is true.
By the principle of mathematical induction, for any non-negative integer $ n $ , $ \cos x \cos (2x) \cos(4x) ... \cos(2^n x)=\frac{\sin(2^{n+1}x)}{2^{n+1}\sin x} $
(b) $ \frac{d}{dx} \left[\frac{1}{2^{101}}\sin(2^{101}x)\csc x \right] $
$ =-\frac{1}{2^{101}}\sin(2^{101}x)\csc x\cot x+\frac{1}{2^{101}}\csc x \cos(2^{101}x)\cdot 2^{101} $
$ =-\frac{1}{2^{101}}\sin(2^{101}x)\csc x\cot x+\csc x \cos(2^{101}x) $
$ \frac{d}{dx}|_{x=\frac{\pi}{4}} \frac{1}{2^{101}}\sin(2^{101}x)\csc x $
$ =-\frac{1}{2^{101}}\sin(2^{101}\cdot \frac{\pi}{4})\csc \frac{\pi}{4}\cot \frac{\pi}{4}+\csc \frac{\pi}{4} \cos(2^{101}\cdot \frac{\pi}{4}) $
$ =\sqrt 2 $
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