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I need help with the following: Use mathematical induction to prove that for every $n\in N$, $$ \sum_{k=1}^n\frac{1}{\cos kx \cos(k+1)x}=\frac{\tan(n+1)x-\tan x}{\sin x} $$ For $n=1$, the statement is true. Suppose that the statement is true for $n=m\in N$, and prove that it is true for $n=m+1$.

$$\sum_{k=1}^{m+1} \frac{1}{\cos kx \cos(k+1)x}=\frac{\tan(m+2)x-\tan x}{\sin x}(*)$$ Proof: $$\frac{\tan(m+2)x-\tan x}{\sin x}=\frac{\tan(m+1)x-\tan x}{\sin x}+\frac{1}{\cos(k+1)x \cos(k+2)x}$$ Using trigonometry rule $\cos x \cos y$, the right side of equation is $$\frac{(\cos x+\cos(2k+3)x)(\tan(m+1)x-\tan x)+2\sin x}{\sin x(\cos x+\cos(2k+3)x)}$$ How to further transform this expression to become $(*)$

Thanks for replies.

wythagoras
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user300045
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2 Answers2

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$$\frac{\tan(m+2)x-\tan x}{\sin x}-\frac{\tan(m+1)x-\tan x}{\sin x}$$

$$=\frac{\tan(m+2)x-\tan(m+1)x}{\sin x}$$

$$=\dfrac{\sin[(m+2)x-\sin(m+1)x]}{\cos(m+2)x\cdot\cos(m+1)x\cdot \sin x}$$

$$=\dfrac1{\cos(m+2)x\cdot\cos(m+1)x}$$ if $\sin x\ne0$

lab bhattacharjee
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this is a telescoping series. you have $$\frac{\sin (k+1)x}{cos(k+1)x} - \frac{\sin kx}{\cos kx}=\frac{\sin( k+1)x\cos kx - \sin kx\cos(k+1)x}{cos(k+1)\cos kx} = \frac{\sin x}{cos(k+1)\cos kx} \tag 1$$

adding the equations for $k = 1, 2, \cdots, n$ gives you $$\frac{\sin (n+1)x}{cos(n+1)x} - \frac{\sin x}{\cos x} = \sin x \sum_{k=1}^n\frac{1}{coskxcos(k+1)x}$$

abel
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