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Mathematical Induction Involving Trigonometry (Compound angles)

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Jefferson :
Hi all,
Sorry for the post spam,
Could someone help me with the questions (in attachment) below.

For:
i.  Is there a quicker or more intuitive approach for this part of the question? (It was shortened in the attachment to fit the page).
I feel like the way I did it isn't the convention.
EDIT: Mistake in the ====> when I was re-copying my answers for conciseness, meant to be tan(α - β) on the denominator. Thanks for pointing that out, supR.

ii. In steps 1 and 3, I feel like I've wasted many lines and used more properties than needed :(. Were there things that I could've simplified to quickened the process?

Thank you so much for your time :'(.

supR:
Maybe this wasn't the answer you were looking for, but it looks good to me. For part iii I would've done it exactly the same, and honestly, showing all the steps is probably safer than trying to skip them and ending up not being able to solve the question. At the end of the day, induction questions are rather secure marks, so it's important to nail the 3-4 marks that are allocated for it. c:

For part i, I believe you wrote the very very first line (with the ==>) wrong, as the denominator should be different? For the method though, I think it's rather smart and I'm not sure how else you'd do it, although I have seen this question before and can't exactly remember how I did it cx

Best of luckk c:

RuiAce:
Just for the record, the intended approach for i) was to make a substitution, which was what you did. Substitution can still sometimes be the most intuitive method.

Don't worry too much about "wasting more lines" - you can always ask for more writing paper in the exam room.

Jefferson :

--- Quote from: RuiAce on January 31, 2019, 08:37:25 pm ---Just for the record, the intended approach for i) was to make a substitution, which was what you did. Substitution can still sometimes be the most intuitive method.


--- End quote ---

Hi RuiAce,
I've just discovered that if you divide the LHS by itself (then dividing the RHS by the LHS), proving with the given property for the RHS = 1 is relatively simpler.
However, I'm not sure if this is acceptable in the proof, since I'm messing with both sides (changing their values) as though they are equations. Is there a way to incorporate or utilise this for questions that requires proving?

RuiAce:

--- Quote from: Jefferson  on January 31, 2019, 09:01:58 pm ---Hi RuiAce,
I've just discovered that if you divide the LHS by itself (then dividing the RHS by the LHS), proving with the given property for the RHS = 1 is relatively simpler.
However, I'm not sure if this is acceptable in the proof, since I'm messing with both sides (changing their values) as though they are equations. Is there a way to incorporate or utilise this for questions that requires proving?

--- End quote ---
It's ultimately still a substitution. The only way you can really formalise it is to say "let \( \alpha = n\theta\) and \(\beta = (n+1)\theta\)" or something like that.
\[ \text{Although once you prove that }\frac{\cot\theta (\tan (n+1)\theta -\tan\theta )}{1+\tan n\theta \tan (n+1)\theta}=1\\ \text{I would still say something like}\\ \therefore1+\tan n\theta\tan(n+1)\theta = \cot\theta(\tan(n+1)\theta-\tan\theta)  \]

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