Mathematical Induction (1 Viewer)

lyounamu · Jul 29, 2008

(ii) Use mathematical induction to prove that, for all integers [FONT=EDLGK F+ Times,Times]n ≥[/FONT]1, tanθ tan 2θ+tan 2θ tan 3θ+[FONT=EDLGL I+ MT] +[/FONT]tan [FONT=EDLGK F+ Times,Times]nθ [/FONT]tan([FONT=EDLGK F+ Times,Times]n +[/FONT]1)θ=−([FONT=EDLGK F+ Times,Times]n +[/FONT]1))+cotθ tan([FONT=EDLGK F+ Times,Times]n +[/FONT]1)θ.

Thi question is ridiculous in my opinion.

You can not even prove that when n=1, the both side of the equation satisfies.

So, I missed the first step and went onto the 2nd and 3rd steps and solved it. But I don't know how to prove that the equation is correct when n = 1. (i.e. LHS = RHS)

tacogym27101990 · Jul 29, 2008

lyounamu said:
(ii) Use mathematical induction to prove that, for all integers [FONT=EDLGK F+ Times,Times]n [/FONT][FONT=EDLGK F+ Times,Times]≥[/FONT]1, tanθ tan 2θ+tan 2θ tan 3θ+[FONT=EDLGL I+ MT] +[/FONT]tan [FONT=EDLGK F+ Times,Times]n[/FONT][FONT=EDLGK F+ Times,Times]θ [/FONT]tan([FONT=EDLGK F+ Times,Times]n [/FONT][FONT=EDLGK F+ Times,Times]+[/FONT]1)θ=−([FONT=EDLGK F+ Times,Times]n [/FONT][FONT=EDLGK F+ Times,Times]+[/FONT]1))+cotθ tan([FONT=EDLGK F+ Times,Times]n [/FONT][FONT=EDLGK F+ Times,Times]+[/FONT]1)θ.

Thi question is ridiculous in my opinion.

You can not even prove that when n=1, the both side of the equation satisfies.

So, I missed the first step and went onto the 2nd and 3rd steps and solved it. But I don't know how to prove that the equation is correct when n = 1. (i.e. LHS = RHS)

is it missing stuff?
like where theres the 2 pluses and the unpaired parentheses???

but yeah this question looks ridiculous
is step 3 hard to prove??

lolokay · Jul 29, 2008

you want to show that tanxtan2x = -(2) + [cotx][tan2x], yeah?

= (2tan^2 x + 2 - 2)/(1 - tan^2 x)
= -2 + 2cotx tanx/(1 - tan^2 x)
= -2 + cotx tan2x

lyounamu · Jul 29, 2008

tacogym27101990 said:
is it missing stuff?
like where theres the 2 pluses and the unpaired parentheses???

but yeah this question looks ridiculous
is step 3 hard to prove??

It wasn't that hard to prove. But I cannot even prove that n=1 from LHS and RHS in the first place!

lyounamu · Jul 29, 2008

lolokay said:
you want to show that tanxtan2x = -(2) + [cotx][tan2x], yeah?

= (2tan^2 x + 2 - 2)/(1 - tan^2 x)
= -2 + 2cotx tanx/(1 - tan^2 x)
= -2 + cotx tan2x

Ah. Yes. I get it now. Thanks.

ProGT408 · Jan 14, 2024

Gonna reopen this thread after a while lol, so how do you prove the second and third steps?

carrotsss · Jan 14, 2024

ProGT408 said:
Gonna reopen this thread after a while lol, so how do you prove the second and third steps?

what’s the actual question because the OP is like unreadable

ProGT408 · Jan 14, 2024

carrotsss said:
what’s the actual question because the OP is like unreadable

yea my bad,

tanθtan2θ+tan2θtan3θ+...tannθtan(n+1)θ=tan(n+1)θcotθ-(n+1)

I got the base case, but after idek how to sub in the inductive step

its worth mentioning there is a part a which gives the identity tana + tanb = tan(a+b) * (1-tanatanb)

WeiWeiMan · Jan 14, 2024

ProGT408 said:
yea my bad,

tanθtan2θ+tan2θtan3θ+...tannθtan(n+1)θ=tan(n+1)θcotθ-(n+1)

I got the base case, but after idek how to sub in the inductive step

its worth mentioning there is a part a which gives the identity tana + tanb = tan(a+b) * (1-tanatanb)

You could try a telescoping sum and see what happens

Luukas.2 · Jan 15, 2024

Here is a complete proof, though there may be simplifications available...

Theorem: For all integers $n \ge 1$

$\sum_{r=1}^n \tan{r\theta}\tan{(r+1)\theta} = \tan{(n + 1)\theta}\cot{\theta} - (n + 1)$

Proof: By induction on $n$ :

A: Put $n = 1:$

$\begin{align*} \text{LHS} &= \tan{\theta}\tan{2\theta} \\ \\ \text{RHS} &= \tan{2\theta}\cot{\theta} - 2 \\ &= \tan{2\theta}\left(\frac{1}{\tan{\theta}} - \frac{2}{\tan{2\theta}}\right) \qquad \text{as $\cot{\theta} = \frac{1}{\tan{\theta}}$} \\ &= \tan{2\theta}\left(\frac{1}{\tan{\theta}} - \frac{2\left(1-\tan^2{\theta}\right)}{2\tan{\theta}}\right) \qquad \text{as $\tan{2\theta} = \frac{2\tan{\theta}}{1-\tan^2{\theta}}$} \\ &= \frac{\tan{2\theta}}{\tan{\theta}}\left(1 - 1 + \tan^2{\theta}\right) \\ &= \frac{\tan{2\theta}\tan^2{\theta}}{\tan{\theta}} \\ &= \tan{\theta}\tan{2\theta} \\ &= \text{LHS} \end{align*}$

Note: Using the identity quoted by @ProGT408, $\color{blue} \tan{a} + \tan{b} = \tan{(a + b)} \times (1 - \tan{a}\tan{b})$ with $\color{blue} a = b = \theta$ gives $\color{blue} 2\tan{\theta} = \tan{2\theta}\left(1 - \tan^2{\theta}\right)$ , which allows:

$\color{blue} \begin{align*} \text{RHS} &= \tan{2\theta}\cot{\theta} - 2 \\ &= \frac{\tan{2\theta}}{\tan{\theta}} - 2 \qquad \text{as $\cot{\theta} = \frac{1}{\tan{\theta}}$} \\ &= \frac{2\tan{2\theta}}{\tan{2\theta}\left(1 - \tan^2{\theta}\right)} - 2 \\ &= \frac{2 - 2\left(1 - \tan^2{\theta}\right)}{\left(1 - \tan^2{\theta}\right)} \\ &= \frac{2\tan^2{\theta}}{\left(1 - \tan^2{\theta}\right)} \\ &= \tan{\theta} \times \frac{2\tan{\theta}}{\left(1 - \tan^2{\theta}\right)} \\ &= \tan{\theta} \times \tan{2\theta} \qquad \text{as $\tan{2\theta} = \frac{2\tan{\theta}}{1-\tan^2{\theta}}$} \\ &= \text{LHS} \end{align*}$

By either method, the result is true when $n = 1$ .

B: Let $k$ be a value of $n$ for which the result is true. That is:

$\sum_{r=1}^k \tan{r\theta}\tan{(r+1)\theta} = \tan{(k + 1)\theta}\cot{\theta} - (k + 1) \qquad \qquad \text{(**)}$

We must now prove the result for $n = k + 1$ . That is, we must prove that

$\sum_{r=1}^{k+1} \tan{r\theta}\tan{(r+1)\theta} = \tan{(k + 2)\theta}\cot{\theta} - (k + 2)$

$\begin{align*} \text{LHS} &= \sum_{r=1}^{k+1} \tan{r\theta}\tan{(r+1)\theta} \\ &= \sum_{r=1}^k \tan{r\theta}\tan{(r+1)\theta} + \tan{(k + 1)\theta}\tan{(k+2)\theta} \\ &= \tan{(k + 1)\theta}\cot{\theta} - (k + 1) + \tan{(k + 1)\theta}\tan{(k+2)\theta} \qquad \text{using the induction hypothesis (**)} \\ &= \tan{(k + 1)\theta}\cot{\theta}\left[1 + \tan{\theta}\tan{(k+2)\theta}\right] - (k + 1) \qquad \text{as $\cot{\theta} = \frac{1}{\tan{\theta}}$} \qquad \qquad \text{. . . (1)} \end{align*}$

Now, considering the given identity:

$\color{blue} \begin{align*} \tan{a} + \tan{b} &= \tan{(a + b)} \times (1 - \tan{a}\tan{b}) \\ \text{Take $a = \theta$ and $b = (k + 1)\theta$:} \qquad \tan{\theta} + \tan{(k + 1)\theta} &= \tan{\left[\theta + (k + 1)\theta\right]} \times \left[1 - \tan{\theta}\tan{(k + 1)\theta}\right] \\ \frac{\tan{\theta} + \tan{(k + 1)\theta}}{\tan{(k + 2)\theta}} &= 1 - \tan{\theta}\tan{(k + 1)\theta} \\ \frac{\tan{\theta} + \tan{(k + 1)\theta}}{\tan{(k + 1)\theta}} &= \frac{\tan{(k +2)\theta}}{\tan{(k+1)\theta}} - \tan{\theta}\tan{(k + 2)\theta} \qquad \text{on multiplying by $\frac{\tan{(k +2)\theta}}{\tan{(k+1)\theta}}$} \\ \tan{\theta}\tan{(k + 2)\theta} &= \frac{\tan{(k +2)\theta}}{\tan{(k+1)\theta}} - \left(\frac{\tan{\theta}}{\tan{(k + 1)\theta}} + 1\right) \\ 1 + \tan{\theta}\tan{(k + 2)\theta} &= \frac{\tan{(k +2)\theta} - \tan{\theta}}{\tan{(k + 1)\theta}} \qquad \qquad \text{. . . (*)} \end{align*}$

$\begin{align*} \text{Returning to (1), we have LHS} &= \tan{(k + 1)\theta}\cot{\theta}\left[1 + \tan{\theta}\tan{(k+2)\theta}\right] - (k + 1) \\ &= \tan{(k + 1)\theta}\cot{\theta} \times \frac{\tan{(k +2)\theta} - \tan{\theta}}{\tan{(k + 1)\theta}} - (k + 1) \qquad \text{using (*)} \\ &= \cot{\theta} \times \left[\tan{(k +2)\theta} - \tan{\theta}\right] - (k + 1) \\ &= \tan{(k +2)\theta}\cot{\theta} - \tan{\theta}\cot{\theta} - (k + 1) \\ &= \tan{(k +2)\theta}\cot{\theta} - (k + 2) \qquad \text{as $\tan{\theta}\cot{\theta} = 1$} \\ &= \text{RHS} \end{align*}$

So, if the result is true for $n = k$ , then it must also be true for $n = k+1$ .

C: It follows from A and B by the process of mathematical induction that the result is true for all positive integers $n$ .

liamkk112 · Jan 15, 2024

heres my proof (excluding base case)

our assumption is that tanxtan2x + tan2xtan3x +... + tankxtan(k+1)x = tan(k+1)x cotx - (n+1) (im using x instead of theta its easier to type)

then we are proving that tanxtan2x + tan2xtan3x +... + tankxtan(k+1)x + tan(k+1)xtan(k+2)x = tan(k+2)xcotx - (k+2)

subbing in our assumption we get

LHS = tan(k+1)xcotx +tan(k+1)xtan(k+2)x - (k+1)

using tan angle sum we see that tan(k+2)x = (tan(k+1)x +tanx)/(1-tan(k+1)xtanx)
so LHS = tan(k+1)x/tanx + tan(k+1)x (tan(k+1)x +tanx)/(1-tan(k+1)xtanx) (cotx = 1/tanx) - (k+1)
then factoring tan(k+1)x and adding fractions:

LHS = tan(k+1)x (1 - tanxtan(k+1)x + tanxtan(k+1)x + tan^2x)/(tanx(1-tanxtan(k+1)x)) - (k+1)
= tan(k+1)x (1+tan^2x)/(tanx(1-tanxtan(k+1)x)) (k+1)
then splitting up the fraction, we now want to recover tan(k+2)x, so we should add tanx and subtract tanx:

LHS =(tan(k+1)x + tanx - tanx)/(tanx(1-tanxtan(k+1)x)) + (tan(k+1)tan^2x)/(tanx(1-tanxtan(k+1)x)) - (k+1)

= tan(k+2)x/tanx - tanx/(tanx(1-tanxtan(k+1)x)) + (tan(k+1)tan^2x)/(tanx(1-tanxtan(k+1)x)) - (k+1) by angle sum
then factoring out tanx we get

LHS = tan(k+2)x/tanx - tanx( 1 - tanxtan(k+1)x)/(tanx(1-tanxtan(k+1)x)) - (k+1)
which cancels to give

LHS = tan(k+2)x/tanx - 1 - (k+1)

then since 1/tanx = cotx and -1 - (k+1) = -(k+2) we get

LHS = tan(k+2)xcotx - (k+2) = RHS as required

Luukas.2 · Jan 15, 2024

Note: The result that I derived in blue, that

$\color{blue} 1 + \tan{\theta}\tan{(k + 2)\theta} = \frac{\tan{(k +2)\theta} - \tan{\theta}}{\tan{(k + 1)\theta}}$

can actually be derived more easily by:

$\color{blue} \begin{align*} \tan{(k + 1)\theta} &= \tan{\left[(k + 2)\theta - \theta\right]} \\ \\ &= \frac{\tan{(k + 2)\theta} - \tan{\theta}}{1 + \tan{(k + 2)\theta}\tan{\theta}} \\ \\ 1 + \tan{\theta}\tan{(k + 2)\theta} &= \frac{\tan{(k +2)\theta} - \tan{\theta}}{\tan{(k + 1)\theta}} \end{align*}$

ProGT408 · Jan 15, 2024

thank you so much guys!

Luukas.2 · Jan 15, 2024

FYI, this a variation on the 2006 MX1 HSC question 5(d).

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Mathematical Induction (1 Viewer)

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