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HELP! Maths C3

swellow5

Simplify cot (pi +x)

Reply 1

metrize

1/tan(pi+x) and use compound angle

Reply 2

swellow5

Explain please, step by step

Posted from TSR Mobile

Reply 3

L33t

Original post by swellow5

Explain please, step by step

Posted from TSR Mobile

Reciprocate (flip it upside down) the double angle formula for tan (this comes from 1/tan(pi+x))using the values in your OP it will simplify when you do this. It is better for you to be given hints and do it yourself otherwise you won't learn. HINT: tan(pi)=0 in radians.

(edited 7 years ago)

Reply 4

swellow5

Oh I see, but what if it was cot(pi - x)

Posted from TSR Mobile

Reply 5

IrrationalRoot

Original post by L33t

Probably better to just understand the fact that

\tan(\pi+x)=\tan x

instead of mechanically resorting to the double angle formula. It's one of the most important properties of the function (

\pi

is its period) and it's very easy to see with the unit circle. It's always best to use the unit circle as much as possible since this enforces understanding of the trig functions as opposed to memorisation of formulae.
But yeah your method is still valid (although strictly speaking

\cot x \not\equiv \dfrac{1}{\tan x}

Reply 6

L33t

Original post by IrrationalRoot

Probably better to just understand the fact that

\tan(\pi+x)=\tan x

instead of mechanically resorting to the double angle formula. It's one of the most important properties of the function (

\pi

\cot x \not\equiv \dfrac{1}{\tan x}

Either method gets you the right answer though, doesn't it? Twas the way I was taught to do it anyhow.

I don't mind either way, I've finished my maths A-Level now thank god!

(edited 7 years ago)

Reply 7

IrrationalRoot

Original post by L33t

Either method gets you the right answer though, doesn't it?

I don't mind either way, I've finished my maths A-Level now thank god!

Right answer for this question, but with that mentality one could struggle with a lot of other questions due to poor understanding of how the functions work.

Reply 8

L33t

Original post by IrrationalRoot

Right answer for this question, but with that mentality one could struggle with a lot of other questions due to poor understanding of how the functions work.

Why is everything on TSR a competition all of a sudden? I was never taught the method you used (edexcel board maths), not that that's an excuse for my apparent sloppy mathematical methods, but sorry my mathematical standards don't live up to your expectations. In future I guess I must try harder.

It is my understanding that strictly speaking

\cot x \not\equiv \dfrac{1}{\tan x}

is beyond the scope of C3 understanding?

Have a nice day :smile:

(edited 7 years ago)

Reply 9

L33t

Original post by swellow5

Oh I see, but what if it was cot(pi - x)

Posted from TSR Mobile

I'd use the same method with the tan(A-B) identity although apparently my methods are inferior!

Reply 10

IrrationalRoot

Original post by L33t

\cot x \not\equiv \dfrac{1}{\tan x}

is beyond the scope of C3 understanding?

Have a nice day :smile:

Not sure why you're getting so wound up; this was mainly for OP's benefit than yours.
Neither am I sure why helping people to improve their mathematical understanding would be considered a 'competition'.

Reply 11

IrrationalRoot

Original post by L33t

I'd use the same method with the tan(A-B) identity although apparently my methods are inferior!

I've just realised you probably misunderstood what I meant when I said this:

"Right answer for this question, but with that mentality one could struggle with a lot of other questions due to poor understanding of how the functions work."

When I said 'with that mentality' I didn't mean the method which you used (which is perfectly valid), I meant the mentality of 'if it gets the right answer, it's fine and you should move on.' Sometimes getting the right answer alone isn't ideal, for the reasons I mentioned before.

Reply 12

math42

tan (and, by extension, cot) has period of pi. i.e. tanx = tan(x - pi) = tan(x + pi) = tan(x + k*pi) for any integer k. This is obvious from the graph.

Reply 13

L33t

Original post by IrrationalRoot

Fair dos, i'm not getting wound up, I genuinely misunderstood what you said then. I thought you were saying I was stupid for using that method and didn't understand what I was saying. I recognise that understanding the trig functions and the domains in which they are repeated is a better and quicker way to deal this question.