Trigonometry, Prove the Identity and more

[rocomath]

you know tan(pi/2) = UD
but cot(pi/2) = 0

so, take cot of both sides
and you can write it as
1/tan(all side)

well wouldn't i need the sum and difference identity for cot?

i proofed it the way i got the sum/diff for tan and got 0 = 0

EDIT: crap, i messed up, let me re-work this ... my handwriting is very bad, lol

by luck i arrived at the same answer, 0 in my numerator.

 

[rootX]

well wouldn't i need the sum and difference identity for cot?

no, you don't
$$cot\left( atan\left( x\right) +acot\left( x\right) \right) =0$$
$$\frac{1}{tan\left( atan\left( x\right) +acot\left( x\right) \right) }=0$$

 

[Gib Z]

$$\tan (\pi/2 -x) = \cot x$$.

 

[rocomath]

my teacher just showed me how to work 52, he did it in less than 5 steps ... amazing

 

[rootX]

he squared them, or used that assumption?

 

[cristo]

my teacher just showed me how to work 52, he did it in less than 5 steps ... amazing

I know I'm jumping into a thread I've probably not read properly here, but in question 52 it says that "I_1 can equal I_2 in some situations in mechanics. Assuming this is the case, simplify." Therefore, one simply sets I_1=I_2, which happens to simplify the equation a great deal.

 

[P.O.L.A.R]

here is a good one prove tan(1/n) = tan(1/[n+1])+tan(1/[n^2+1])

note: I don't have the problem with me I have make sure it is addition which I am ppositive it is.

 

[matadorqk]

1=2

Nope, otherwise you could prove that 1=2.

Which is clearly true.

 

[rootX]

here is a good one prove tan(1/n) = tan(1/[n+1])+tan(1/[n^2+1])

note: I don't have the problem with me I have make sure it is addition which I am ppositive it is.

after, spending an hour, I plugged in some random number:
$$f(n) = -tan\left( [\frac{1}{{n}^{2}+1}]\right) -tan\left( [\frac{1}{n+1}]\right) +tan\left( \frac{1}{n}\right)$$

f(6)
$$-tan\left( [\frac{1}{7}]\right) -tan\left( [\frac{1}{37}]\right) +tan\left( \frac{1}{6}\right)$$

I guess that really doesn't give a 0!

 

[P.O.L.A.R]

after, spending an hour, I plugged in some random number:
$$f(n) = -tan\left( [\frac{1}{{n}^{2}+1}]\right) -tan\left( [\frac{1}{n+1}]\right) +tan\left( \frac{1}{n}\right)$$

f(6)
$$-tan\left( [\frac{1}{7}]\right) -tan\left( [\frac{1}{37}]\right) +tan\left( \frac{1}{6}\right)$$

I guess that really doesn't give a 0!

You have to remember that when working with proofs you can only work one side of the proof not both. Usually it is best to start with the side that has the most variables.

 

[Gib Z]

If rootX is correct is that his expression doesn't not give zero, his *proof* is always a disproof by form of the counterexample n=6.

 

[P.O.L.A.R]

here is a good one prove tan(1/n) = tan(1/[n+1])+tan(1/[n^2+1])

note: I don't have the problem with me I have make sure it is addition which I am ppositive it is.

arctan(1/n) = arctan(1/[n+1])+arctan(1/[n^2+n+1])

forgot the n sorry. also I thought I put arctan but I guess not wow I really messed that one up I aologize. A little tired I guess.

 

[dextercioby]

Apply tangent on both sides of the equation. What do you get ?

 

[P.O.L.A.R]

Apply tangent on both sides of the equation. What do you get ?

I already proved it myself people were asking proof questions and I found this one to be a good one.

Starting that way didnt work for me. I couldn't figure out where to go after taking the tan.