Math Is Fun Forum

Whizzies

Member

Registered: 2014-07-18

Posts: 53

Trigonometry

Hello everybody!

I am stuck with these functions:

It is about trignometry the functions of sine and cosine;
This is the formula for line symmetry: f(a-p)=f(a+p) I can follow the reasoning behind that.

but point symmetry just blows my mind:

f(a-p)-b=b-f(a+p)
Left side and right side

I don't understand it! please help me I can't find this on the net.

Bob

Administrator

Registered: 2010-06-20

Posts: 10,621

Re: Trigonometry

hi Whizzies,

I've not seen this way of defining point symmetry before so I've had to do a bit of experimenting.  When I come across a tricky bit of math, I try to simplify the numbers first, and also pick an example for what I'm trying to investigate.

For symmetry, I wouldn't choose a trig. function because the graphs have too many lines and points of symmetry, so its impossible to see what's going on.

I chose y = x^3, because it only has point symmetry around (0,0)

You'll find it is worth using the function grapher at

http://www.mathsisfun.com/data/function … c1=(1+x)^3

I've set this up with the function y = (1+x)^3.  Try some values:

x = -3 ,    y = -8
x = -2 ,    y = -1
x = -1 ,    y = 0
x = 0  ,    y = 1
x = 1  ,    y = 8

You can see from this that the graph had point symmetry around (-1,0)

Try changing that '1' into other numbers and you'll see that, with a = -1

And f(a-x) = -f(a+x)

Now let's try adding a constant.

y = x^3 + 2

You'll see that this shifts the graph up by 2, so the symmetry point is moved to (0,2)

Then

Here I've put square brackets around the function and introduced an additional '2' (either + 2 or -2) to balance the equation

so in general

You don't need the square brackets here, but I thought it would help you to compare with the line above if I left them in.

This function as point symmetry around (0,b)

Finally put both bits together:

So y = (1+x)^3 + 2

x = -3 , y = -6
x = -2 , y = 1
x = -1 , y = 2
x = 0 , y = 3
x = 1 , y = 10

Try it on the grapher as well.

You'll see the point symmetry is now around (-1,2)

Again you can vary the value of the constant.

So if I put in square brackets for the function again:

y = [(1+x)^3 + 2]

a = -1 and b = 2

and

so in general

with the point of symmetry (-a,b)

Bob

Last edited by Bob (2014-11-21 01:00:57)

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Children are not defined by school ...........The Fonz
You cannot teach a man anything;  you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you!  …………….Bob

Whizzies

Member

Registered: 2014-07-18

Posts: 53

Re: Trigonometry

okay! I think it is best if I just learn the formulas for this, because I see it is more difficult than anticipated. I can follow your reasoning, but I cannot understand it in the sinus graph and sadly that is the way I wanted to see it

I am now practising a lot with the sine and unit circle to understand it properly. I even found some cool stuff online that will help me out.

Thanks thought for explaining this to me.