1. why is cos(-4x)=cos(4x)?

why is cos(-4x)=cos(4x)? teach me concepts. can add on to sin(-4x) tan(-4x)

2. Originally Posted by stupidguy
why is cos(-4x)=cos(4x)? teach me concepts. can add on to sin(-4x) tan(-4x)
Its because cos(x) is an even function (symmetric about the y axis), meaning that $\displaystyle f(-x)=f(x)$. Now, sin(x) is an odd function (symmetric about the origin), meaning $\displaystyle f(-x)=-f(x)$. You can see why this is the case by looking at the graphs of these two functions.

Now, tan(x) is an odd function, since $\displaystyle \tan (-x)=\dfrac{\sin (-x)}{\cos(-x)}=\dfrac{-\sin x}{\cos x}=-\tan x$.

Does this clarify things?

3. Hello, stupidguy!

$\displaystyle \text{Why is }\,\cos(-4x)=\cos(4x)\;?$

We know: .$\displaystyle \cos\theta \:=\:\frac{adj}{hyp}$

Look at the graphs:

Code:
      |
|           *
|  hyp   *  |
|     *     |
|  * 4x     |
- * - - - - - * - -
|    adj

Code:
      |
- * - - - - - * - -
|  * -4x    |
|     *     |
|   hyp  *  |
|           *
|

Get it?

4. Originally Posted by Soroban
Hello, stupidguy!

We know: .$\displaystyle \cos\theta \:=\:\frac{adj}{hyp}$

Look at the graphs:

Code:
      |
|           *
|  hyp   *  |
|     *     |
|  * 4x     |
- * - - - - - * - -
|    adj

Code:
      |
- * - - - - - * - -
|  * -4x    |
|     *     |
|   hyp  *  |
|           *
|

Get it?

I dun get u!

5. Originally Posted by stupidguy
I dun get u!
In the first one:
adjacent is positive, hypotenuse is positive. cos of angle 4x becomes +adj/+hyp = +ve.

In the second one:
adjacent is positive, hypotenuse is positive. cos of angle -4x remains +adj/+hyp = +ve.

6. Originally Posted by Unknown008
In the first one:
adjacent is positive, hypotenuse is positive. cos of angle 4x becomes +adj/+hyp = +ve.

In the second one:
adjacent is positive, hypotenuse is positive. cos of angle -4x remains +adj/+hyp = +ve.
however u r making an assumption that 4x is smaller than 90 degree. what if it is 135 degree?

7. One way to see it would be to find the Taylor series expansion of $\displaystyle f(x) = \cos{x}$ about $\displaystyle x = 0$ and note that all the powers of $\displaystyle x$ are even, but I realise that this is in the pre-Calculus section.
Originally Posted by Soroban
Hello, stupidguy!
And this is why people should sign-up with appropriate names.

8. If $\displaystyle 4x=135^0$, then $\displaystyle -4x=360^o-4x=360^o-135^o=225^o$

$\displaystyle 135^o=180^o-45^o$

$\displaystyle 225^o=180^o+45^o$

hence, both angles have the same horizontal co-ordinate on the unit-circle.

Cos(angle)=horizontal co-ordinate of a point on the unit-circle circumference,
hence it can be located on the horizontal axis.

If $\displaystyle 4x=220^o\Rightarrow\ -4x=360^o-220^o=140^o$

$\displaystyle 220^o=180^o+40^o$

$\displaystyle 140^o=180^o-40^o$

so the angles have the same horizontal co-ordinate.

Think of Cos(angle) as the horizontal co-ordinate of a point on the circle.
Sin(angle) is the vertical co-ordinate.
Tan(angle) is the slope of the line going through the origin (centre of unit-circle) and the point on the circle.
Or Tan(angle) is Sin(angle) divided by Cos(angle).

A positive angle is an anticlockwise movement starting at zero.
A negative angle is a clockwise movement starting at 360 degrees.