# How to get the beginning of a sine function

I have a function $f(x) = \sin (2x- \dfrac{1}{3} \pi)$ on the domain $[0, 1\dfrac{1}{2} \pi ]$

I want to get the beginning of this function. How do I to this?

I know the x-coordinate is $\dfrac{1}{6}\pi$, but I tried getting the y-coordinate by putting x = 0 but I got a value very close to 0 (which is the correct answer)

Also, my textbook asked for the beginning of the function $g(x) = -\cos (x + \dfrac{1}{6} \pi$)

I tried the y-coordinate, which is almost equal to -1 ( $-cos(\dfrac{1}{6} \pi$)

But when I tried the x-coordinate, this is what I did:

$-\cos (x+ \dfrac{1}{6} \pi) = 0$

$x + \dfrac{1}{6}\pi = 90$, why is this incorrect?

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What is the beginning of a function? –  Ayman Hourieh Mar 8 '13 at 21:55
What is the "beginning" of a function? –  Arjang Mar 8 '13 at 21:58
@AymanHourieh the beginning –  Ylyk Coitus Mar 8 '13 at 22:01
@Arjang the beginning –  Ylyk Coitus Mar 8 '13 at 22:03
Just repeating what you said doesn't help. It sounds like the beginning is the value at the left endpoint of the interval, here $0$. Is that correct? –  Ross Millikan Mar 8 '13 at 22:07

If you are referring to the value at the left end of the interval, here $0$, you should plug $0$ in for $x$. That would give you $f(0)=\sin(-\frac 13 \pi)$, which is definitely not zero.
Based on the comment, you are looking for a root of the function, a place where the value is zero. If $\sin(y)=0$, we must have $y=k\pi$ for some integer $k$. You can verify that $x=\frac 16 \pi$ satisfies that, so $(\frac 16 \pi,0)$ is on the graph. There are two more points in your interval that do as well: $\frac 46 \pi$ and $\frac 76 \pi$
For your addition, trying to find a root of $-\cos (x+\frac 16 \pi)$, note that $\cos(y)=0$ when $y=\frac \pi2+k\pi$ for $k$ some integer, so you want $x+\frac 16\pi=\frac \pi 2$ (please don't mix degrees in, as it appears you have with your $90$), so $x=\frac \pi3$. This would give $(\frac \pi3,0)$ as a root of the function. Again, $(\frac 43\pi,0)$ is another root in your interval.
But wait: My textbook says the correct answer is $(\dfrac{1}{6}\pi, 0)$.. are they wrong? I noticed it was weird already, but maybe they were approximating –  Ylyk Coitus Mar 8 '13 at 22:11