Mathematical Induction question (1 Viewer)

[steve001]

hey, i got this question out of the excel book and it might have been in a past paper.

its:
(a)Write an expression for tan(A-B) and hence deduce that for any positive interger n:
1+tan(na)tan(n+1)a-(n+1) = (tana)^-1 [tan(n+1)a-tan(na).

the a is supposed to be tetra

(b) Using (a), prove that:
Sn = tanatan2a+tan2atan3a+.......+tan(na)tan(n+1)a
= cotatan(n+1)a-(n+1).

 

[youngminii]

tan(A - B) = (tanA - tanB)/(1 + tanAtanB)

So, statement to prove : 1 + tan(na)tan(n + 1)a - (n + 1) = (tana)^-1 [tan(n + 1)a - tan(na)
Wait what the fuck is tetra? You mean theta?
Dude, set out your question properly

Anyways, test for n = 1
LHS = 1 + tan(na)tan(n + 1)a - (n + 1)
= 1 + tan(a)tan(2a) - 2
= tan(a)tan(2a) - 1

RHS = (tana)^-1 [tan(2) - tan(a)] {I am assuming this is multiplication}
= (tan(2) - tan(a))/tan(a)
= tan(2)/tan(a) - 1

Okay well since they don't match, I'm obviously not reading your question right. Re-write your RHS more clearly please, thanks

 

[steve001]

thats how they set it out in the excel book

 

[youngminii]

thats how they set it out in the excel book

You're a stooge. I can't tell what the right hand side is saying. Is it saying multiply all that? 'Cause it looks like two different numbers, just there side-by-side.

 

[alakazimmy]

(a) tan(A - B) = (tanA - tanB)/(1 + tanAtanB)

You use this to manipulate the equation.
Let A = (n+1)a, and B = (n)a --- a is theta

So:
tan((n+1)a - na) = (tan(n+1)a - tan(na))/(1+tan((n+1)a)*tan(na))

Simplifying:
tan(a) = (tan(n+1)a - tan(na))/(1+tan((n+1)a)*tan(na))
Hence: (1+tan((n+1)a)*tan(na)) = tan(a)^-1 * (tan(n+1)a - tan(na))

No idea where the extra -(n+1) came from, check your textbook.

(b) If we use the result, it is:
(tan((n+1)a)*tan(na) = cot(a)*(tan(n+1)a - tan(na)) - 1

Letting n=1, tan(a)*tan(2a) = cot(a)*(tan(2a)-tan(a)) - 1
n=2, tan(2a)*tan(3a) = cot(a)*(tan(3a)-tan(2a)) - 1
...
...
n=n, tan(na)*tan((n+1)a) = cot(a)*(tan((n+1)a)-tan(na))-1

You should see quite an obvious pattern here.

So, tan(a)*tan(2a)+tan(2a)*tan(3a)+.......+tan(na)tan((n+1)a) = cot(a)*(tan((n+1)a))-cot(a)*tan(a)) - n

But, cot(a)*tan(a) = 1
Therefore, tan(a)*tan(2a)+tan(2a)*tan(3a)+.......+tan(na)tan((n+1)a) = cot(a)*(tan((n+1)a)) - (n+1)

 

[alakazimmy]

Oh yea, I did this question without induction. If you need an induction solution, let me know.

 

[Trebla]

The question itself didn't even say to use induction...

 

[alakazimmy]

The question itself didn't even say to use induction...

Yea, I know. I'm assuming the question might be from the "Mathematical Induction" section of the excel textbook.

 

[3.14159potato26]

(a)Write an expression for tan(A-B) and hence deduce that for any positive interger n:
1+tan(na)tan(n+1)a-(n+1) = (tana)^-1 [tan(n+1)a-tan(na).

What's with all the notation ambiguity? First off, if read conventionally, it would mean (tana) to the power of -1 [tan(n+1)a-tan(na)]. So, is it that or the multiplication of (tana) to the power of -1 with the other term [tan(n+1)a-tan(na)]?
Another thing is the meaning of (tana)^-1. No offense, but that is very confusing. One interpretation is that you mean arctan(a), i.e. the inverse tangent function of a. Another interpretation is that is tan(a) to the power of -1, in which case it would mean the reciprocal of tan(a). Clearly, arctan(a) != (1/tan(a)), so which one is it? Please write your question more clearly.

Edit: The symbol "!=" means "not equal to".