Also,

Why is sin(-x) = -sin(x)?

I'm sure I learned this all before, but I've forgotten.

thanks!

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- Aug 6th 2010, 02:23 AMkselogatrWhy is cosine(-x) equal to cosine(x)?
Also,

Why is sin(-x) = -sin(x)?

I'm sure I learned this all before, but I've forgotten.

thanks! - Aug 6th 2010, 02:33 AMAckbeet
The most direct way to show this, in my opinion, is from the definition of cosine as adjacent over hypotenuse. Think of a circle of radius 1 centered at the origin. Draw a straight line (neither vertical nor horizontal) from the origin to a point on the circle in the first quadrant. Call its angle theta. From that intersection point, draw a vertical line down to the x axis. Now, the length of the x axis from the origin to the perpendicular you drew is the adjacent side, correct? It has to be labeled with a positive number, because x is positive in the first quadrant (so is y, but that's not relevant to the cosine function). The hypotenuse is always positive, because it's a length. Therefore, the cosine is positive. Finally, the angle theta is positive, because you're in the first quadrant.

Now flip your original straight line about the x axis. You now have a negative angle: -theta, which is in the fourth quadrant. The adjacent side is still positive, because x is always positive in the fourth quadrant, and the hypotenuse is always positive. Therefore, the cosine function will still be positive. In addition, its value will be the same as before.

You can repeat this argument in the second and third quadrants, where the cosine function is negative (since x is negative there), to convince yourself that, indeed, the cosine function is even.

A similar argument will show you, by the definition of the sine function, that it is odd.

Alternatively, you can show the result from the Taylor series expansions of the two functions. The cosine function has only even powers of x in its expansion, and the sine function has only odd powers of x in its expansion. The result follows immediately from that.

Does this make sense? - Aug 6th 2010, 02:37 AMProve It
Or simply graph the two functions.

It's pretty easy to see that with $\displaystyle \cos{(x)}$ everything to the left of the $\displaystyle y$ axis is a reflection of everything to the right of the $\displaystyle y$ axis.

So $\displaystyle f(-x) = f(x)$ which means $\displaystyle \cos(-x) = \cos(x)$. - Aug 6th 2010, 08:21 AMGrandad
Hello kselogatr

Welcome to Math Help Forum!As soon as you start to look at the sine and cosine of angles outside the range $\displaystyle 0^o$ to $\displaystyle 90^o$, you need a definition which doesn't rely on the ratios of the sizes of the sides of a right-angled triangle.

Instead, you need to look at the Unit Circle.

If you study the Wikipedia page above, you'll see that the cosine and sine of the angle $\displaystyle \displaystyle t$ are*defined*as the $\displaystyle \displaystyle x$- and $\displaystyle \displaystyle y$- coordinates respectively of the point that moves around the circle in an*anticlockwise*direction, starting at the point $\displaystyle \displaystyle (1,0) $ on the $\displaystyle \displaystyle x$-axis.

A moment's thought will convince you that if the point moves*clockwise*instead (and therefore through*negative*values of $\displaystyle \displaystyle t$) its $\displaystyle \displaystyle x$-coordinate (the cosine) is exactly the same as before.So $\displaystyle \cos (-t) = \cos (t)$.

But its $\displaystyle \displaystyle y$-coordinate is $\displaystyle \displaystyle -1$ times what it was before.So $\displaystyle \sin(-t) = \sin(t)$OK now?

Grandad

- Aug 6th 2010, 05:56 PMkselogatr
Thanks everyone! I didn't expect replies

*this*quickly.