A12c - Graphs of trigonometric functions - BossMaths.com - Free Printable

Let me carefully examine the sine graph and identify all the labeled points. The graph is of $ y = \sin(x) $, with x-values in degrees (since we see -360, -270, ..., 360 on the x-axis).

The points marked are:
- $(c, 0.5)$ — left side, rising part before peak at -270°
- $(-210, d)$ — purple point, left of center, above x-axis
- $(30, 0.5)$ — red point, right side, rising part before peak at 90°
- $(a, 0.5)$ — purple point, right side, after peak at 90°, descending
- $(g, -0.5)$ — green point, left side, descending after trough at -90°
- $(b, -0.5)$ — blue point, near origin, descending toward trough at -90°
- $(210, e)$ — green point, right side, descending after peak at 90°, before trough at 270°
- $(f, -0.5)$ — blue point, right side, ascending after trough at 270°

We need to find the missing coordinates: $c, d, a, g, b, e, f$.

Since this is $y = \sin(x)$, and the graph is symmetric and periodic with period 360°, and we know standard sine values:

- $\sin(30^\circ) = 0.5$
- $\sin(150^\circ) = 0.5$ (since sin is symmetric about 90°: $\sin(180^\circ - x) = \sin x$)
- $\sin(-210^\circ) = \sin(150^\circ)$? Wait: use identity $\sin(-\theta) = -\sin(\theta)$, but better to add 360°: $-210^\circ + 360^\circ = 150^\circ$, so $\sin(-210^\circ) = \sin(150^\circ) = 0.5$. So $d = 0.5$

Wait — check: $\sin(-210^\circ) = \sin(-(180+30)) = -\sin(180+30) = -(-\sin 30) = \sin 30 = 0.5$. Yes, correct.

So point $(-210, d)$ has $d = 0.5$

Now $(c, 0.5)$: same y = 0.5. We already have $(30, 0.5)$ and $(a, 0.5)$. Since sine = 0.5 at angles: $30^\circ + 360k$ and $150^\circ + 360k$ for any integer k.

On the left side (negative x), the next solution before 30° is $30^\circ - 360^\circ = -330^\circ$, and $150^\circ - 360^\circ = -210^\circ$. But $(-210, d)$ is already listed and we just found $d = 0.5$, so that’s one point where y=0.5 at x = -210°.

Then what is $(c, 0.5)$? Look at the graph: the red dot at $(c, 0.5)$ is left of the peak at -270°, i.e., between -360° and -270°, and it's on the rising part (since sine rises from -360° (0) to -270° (1)). So x must be -330°, because:
- $\sin(-330^\circ) = \sin(30^\circ) = 0.5$ (since -330 + 360 = 30)

Yes! So $c = -330$

Next: $(a, 0.5)$ — purple point on right side, after peak at 90°, descending. That should be at $150^\circ$, since $\sin(150^\circ) = 0.5$. So $a = 150$

Now points with y = -0.5:
We know $\sin(x) = -0.5$ at:
- $x = -30^\circ + 360k$
- $x = -150^\circ + 360k$ (since $\sin(210^\circ) = -0.5$, and $210 = 180+30$, also $\sin(330^\circ) = -0.5$)

Standard solutions in [0°, 360°]: $210^\circ$ and $330^\circ$
In [-360°, 0°]: $-150^\circ$ and $-30^\circ$

Check the points:
- $(g, -0.5)$: green point, left side, between -270° and -180°, i.e., in third quadrant → that’s $-150^\circ$? Wait: -150° is between -180° and -90°, not -270° and -180°.

Let’s map intervals:
- From -360° to -270°: sine goes from 0 → 1 (rising)
- -270° to -180°: sine goes from 1 → 0 (falling)
- -180° to -90°: sine goes from 0 → -1 (falling)
- -90° to 0°: sine goes from -1 → 0 (rising)

So point $(g, -0.5)$ is labeled near x ≈ -210? No — look: there’s a green dot labeled $(g, -0.5)$ below the x-axis, between -270° and -180°? Actually, looking at the graph: the green dot $(g, -0.5)$ is between -180° and -90°, because it's after the zero at -180°, going down toward trough at -90°. So that’s in the interval (-180°, -90°), where sine is negative and decreasing.

In that interval, sine = -0.5 at $x = -150^\circ$, because:
$\sin(-150^\circ) = -\sin(150^\circ) = -0.5$. Yes.

So $g = -150$

Next: $(b, -0.5)$ — blue point, between -90° and 0°, rising from -1 to 0. So sine = -0.5 there at $x = -30^\circ$, since $\sin(-30^\circ) = -0.5$. So $b = -30$

Next: $(210, e)$ - green point at x = 210°, we know $\sin(210^\circ) = \sin(180+30) = -\sin(30) = -0.5$, so $e = -0.5$

Finally: $(f, -0.5)$ - blue point on far right, after trough at 270°, rising toward 360°, so in (270°, 360°). Sine = -0.5 there at $x = 330^\circ$, since $\sin(330^\circ) = -\sin(30^\circ) = -0.5$. So $f = 330$

Let me double-check all:

| Point | x | y | Expected sin(x) |
|-------|----|----|----------------|
| (c, 0.5) | -330 | 0.5 | sin(-330)=sin(30)=0.5 ✔ |
| (-210, d) | -210 | ? | sin(-210)=sin(150)=0.5 ⇒ d=0.5 ✔ |
| (30, 0.5) | 30 | 0.5 | sin(30)=0.5 ✔ |
| (a, 0.5) | 150 | 0.5 | sin(150)=0.5 ✔ |
| (g, -0.5) | -150 | -0.5 | sin(-150)=-0.5 ✔ |
| (b, -0.5) | -30 | -0.5 | sin(-30)=-0.5 ✔ |
| (210, e) | 210 | ? | sin(210)=-0.5 ⇒ e = -0.5 ✔ |
| (f, -0.5) | 330 | -0.5 | sin(330)=-0.5 ✔ |

All consistent.

So missing coordinates:
- $c = -330$
- $d = 0.5$
- $a = 150$
- $g = -150$
- $b = -30$
- $e = -0.5$
- $f = 330$

But the question says: “find the missing x- and y-coordinates of the points marked.” So each point has one missing coordinate.

List:
- $(c, 0.5)$ → find c = -330
- $(-210, d)$ → find d = 0.5
- $(30, 0.5)$ → both given (not missing)
- $(a, 0.5)$ → find a = 150
- $(g, -0.5)$ → find g = -150
- $(b, -0.5)$ → find b = -30
- $(210, e)$ → find e = -0.5
- $(f, -0.5)$ → find f = 330

Thus final answer should list all missing values.

Since the instruction says: “Final Answer: [Provide only the final correct answer.]”, and no other sections requested, I will give the missing coordinates clearly.

I’ll format as:
c = -330, d = 0.5, a = 150, g = -150, b = -30, e = -0.5, f = 330

But to be safe, maybe list as ordered pairs with missing values filled.

However, the problem likely expects just the values. Let me output cleanly:

Final Answer:
c = -330, d = 0.5, a = 150, g = -150, b = -30, e = -0.5, f = 330