Trig equation (1 Viewer)

Mahan1 · Feb 19, 2017

Prove for $n\geq 2$

$tan(\alpha) \ tan(2\alpha)+ tan(2\alpha) \ tan(3\alpha)+ ...+ tan((n-1)\alpha) \ tan(n\alpha) = \frac{tan(n\alpha)}{tan(\alpha)}-n$

Mahan1 · Feb 26, 2017

well, the solution I came up with relies on induction, but I believe there is another way to solve it as well.
Proof:
prove the base case:
n=2

$LHS = tan(\alpha) \ tan(2\alpha) = tan(\alpha)*\frac{2\ tan(\alpha)}{1-tan^{2}(\alpha)} = \frac{2tan^{2}(\alpha)}{1-tan^{2}(\alpha)}$

$= -\frac{2(1-tan^{2}(\alpha))+2}{1-tan^{2}(\alpha)} = -2 + \frac{tan(2 \alpha)}{tan(alpha)} = LHS$

Suppose it is true for n=k:

$tan(\alpha)\ tan(2\ \alpha) +....+tan((k-1)\ \alpha)\ tan(k\ \alpha) = \frac{tan(k \ \alpha)}{tan{\alpha}} - k \ \ (*)$

we need to prove n=k+1:

$tan(\alpha)\ tan(2\ \alpha) +....+tan(k\ \alpha)\ tan((k+1)\ \alpha) = \frac{tan((k+1) \ \alpha)}{tan{\alpha}} - (k+1)$

Adding $tan(k\ \alpha)\ tan((k+1)\ \alpha)$ to (*):

$tan(\alpha)\ tan(2\ \alpha) +....+tan(k\ \alpha)\ tan((k+1)\ \alpha) = (\frac{tan(k \ \alpha)}{tan{\alpha}} - k) +tan(k\ \alpha)\ tan((k+1)\ \alpha)$

$RHS = (\frac{tan(k \ \alpha)}{tan{\alpha}} - k) +tan(k\ \alpha)\ tan((k+1)\ \alpha = \frac{tan(k \ \alpha)}{tan{\alpha}} +1 +tan(k\ \alpha)\ tan((k+1)\ \alpha -(k+1)$

$( \textrm{Note} 1 +tan(k\ \alpha)\ tan((k+1)\ \alpha = \frac{tan((k+1)\ \alpha) - tan(k \ \alpha)}{tan(\alpha)} )$

$= \frac{tan(k \ \alpha)}{tan{\alpha}} +\frac{tan((k+1)\ \alpha) - tan(k \ \alpha)}{tan(\alpha)} - (k+1) = \frac{tan((k+1)\ \alpha)}{tan(\alpha)} - (k+1)$

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Trig equation (1 Viewer)

Mahan1

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Mahan1

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