Pythagoras & Trigonometry - Free Printable

To match the graphs to their equations, we need to look at key features of each graph: the shape (sine, cosine, or tangent), the vertical shift (up or down), and any reflections (flips).

Here is the step-by-step breakdown for each graph:

1. Analyze the Sine and Cosine Graphs (A-H)
These graphs are wave-shaped. We can identify them by where they start on the y-axis (at $x=0$) and their maximum/minimum values.

* Graph A: Starts at $(0,0)$, goes up to a max of $1$, and down to a min of $-1$. This is the standard sine wave.
* Equation: $y = \sin(x)$

* Graph B: Starts at $(0,-3)$, goes up to $-1$, and down to $-3$. The midline is $y=-2$. It looks like a cosine wave shifted down. Let's check the options. We have $y = \cos(x) - 2$. At $x=0$, $\cos(0)-2 = 1-2 = -1$. Wait, Graph B starts at $-3$. Let's re-examine Graph E.
* Let's look at Graph E: Starts at $(0,-1)$, goes down to $-3$, up to $-1$. Midline is $y=-2$. At $x=0$, value is $-1$. This matches $y = \cos(x) - 2$ because $\cos(0)=1$, so $1-2=-1$. At $x=180$, $\cos(180)=-1$, so $-1-2=-3$. This fits Graph E perfectly.
* So, Graph E is $y = \cos(x) - 2$.

* Let's re-evaluate Graph B: Starts at $(0,-3)$? No, looking closely at the grid, the y-axis labels are $4, 3, 2, 1, 0, -1, -2, -3, -4$.
* Graph B starts at $y=-3$? No, it looks like it starts at $y=-3$ and goes to $-1$? Let's look at the amplitude. Max is $-1$, Min is $-3$. Midline is $-2$. Amplitude is $1$.
* Let's check $y = -\cos(x) - 2$? Not an option.
* Let's check $y = \sin(x) - 3$? At $x=0$, $\sin(0)-3 = -3$. At $x=90$, $\sin(90)-3 = 1-3 = -2$. At $x=270$, $\sin(270)-3 = -1-3 = -4$. Graph B has a minimum of $-4$? Looking at Graph B, the lowest point is at $x=270$ and $y=-4$. The highest point is at $x=90$ and $y=-2$. The starting point is $(0,-3)$. This matches $y = \sin(x) - 3$ perfectly.
* So, Graph B is $y = \sin(x) - 3$.

* Graph C: Starts at $(0,1)$, goes down to $-1$, up to $1$. This is a standard cosine wave but reflected? No, standard $\cos(x)$ starts at $1$, goes to $-1$ at $180$, back to $1$ at $360$. Graph C does exactly this.
* So, Graph C is $y = \cos(x)$.

* Graph D: Starts at $(0,2)$, goes up to $3$, down to $1$. Midline is $y=2$. Amplitude is $1$. It looks like a cosine wave shifted up by $1$? No, shifted up by $2$? If it were $\cos(x)+2$, it would go from $3$ to $1$. Here it starts at $2$. Wait.
* Let's look at the starting point again. Graph D starts at $y=2$. At $x=90$, it is at $y=3$. At $x=180$, it is at $y=2$. At $x=270$, it is at $y=1$. At $x=360$, it is at $y=2$.
* This is a sine wave shifted up by $2$. $\sin(0)+2 = 2$. $\sin(90)+2 = 3$. $\sin(180)+2 = 2$. $\sin(270)+2 = 1$.
* So, Graph D is $y = \sin(x) + 2$.

* Graph F: Starts at $(0,0)$, goes down to $-1$, up to $1$. This is a reflected sine wave. $\sin(x)$ goes up first; $-\sin(x)$ goes down first.
* So, Graph F is $y = -\sin(x)$.

* Graph G: Starts at $(0,-1)$, goes up to $1$, down to $-1$. Midline is $y=0$. Amplitude is $1$. It looks like a negative cosine wave. $-\cos(0) = -1$. $-\cos(90) = 0$. $-\cos(180) = -(-1) = 1$.
* So, Graph G is $y = -\cos(x)$.

* Graph H: Starts at $(0,2)$, goes down to $1$, up to $2$, up to $3$? No.
* Let's trace Graph H carefully. Starts at $y=2$. At $x=90$, $y=1$. At $x=180$, $y=2$. At $x=270$, $y=3$. At $x=360$, $y=2$.
* This behaves like a cosine wave shifted up by $1$? $\cos(0)+1 = 2$. $\cos(90)+1 = 1$. $\cos(180)+1 = 0$. But Graph H is at $2$ at $x=180$.
* Let's re-read the graph. Maybe it's not cosine.
* Let's check $y = \cos(x) + 1$. Max is $2$, Min is $0$. Graph H has Max $3$? No, looking at H, the peak is at $x=270$ with $y=3$? And trough at $x=90$ with $y=1$?
* If it's a sine wave: $\sin(270) = -1$. $-\sin(270) = 1$.
* Let's look at the options remaining for sine/cosine: $y = \cos(x) + 1$.
* Let's test $y = \cos(x) + 1$ on Graph H.
* $x=0 \rightarrow y=2$. (Matches)
* $x=90 \rightarrow y=1$. (Matches)
* $x=180 \rightarrow y=0$. (Graph H shows $y=2$?? No, look closer at H. At $x=180$, the curve is at $y=2$? Or is it passing through?)
* Actually, let's look at Graph H again. It starts at $2$. Goes DOWN to $1$ at $90$. Goes UP to $2$ at $180$? No, that would be a very flat wave.
* Let's look at the shape. It looks like a cosine wave. Standard cosine starts at max. This starts at "mid-height".
* Let's reconsider Graph D and H.
* Graph D: Starts at $2$, goes UP to $3$ at $90$. This is $+\sin(x)$ shifted up. $y=\sin(x)+2$. Correct.
* Graph H: Starts at $2$, goes DOWN to $1$ at $90$. Then back to $2$ at $180$? Then up to $3$ at $270$? Then back to $2$ at $360$?
* If it goes $2 \rightarrow 1 \rightarrow 2 \rightarrow 3 \rightarrow 2$, that is a negative sine wave shifted up by $2$? $y = -\sin(x) + 2$. That is not an option.
* Let's look at the options again. We have $y = \cos(x) + 1$.
* Let's test $y = \cos(x) + 1$:
* $x=0, y=2$.
* $x=90, y=1$.
* $x=180, y=0$.
* $x=270, y=1$.
* $x=360, y=2$.
* Does Graph H show a minimum of $0$ at $x=180$? Looking at the image for H... The curve at $x=180$ is clearly above the x-axis. It looks like it touches $y=1$? No, at $x=90$ it is at $1$. At $x=180$ it is higher.
* Wait, I might have misidentified Graph C.
* Graph C: Starts at $1$, goes to $-1$ at $180$, back to $1$. This is definitely $y=\cos(x)$.
* What about Graph H? Let's look at the remaining equation: $y = \cos(x) + 1$.
* If Graph H is $y = \cos(x) + 1$, the minimum should be $0$. In the drawing for H, the lowest point is at $x=180$. Is that point at $y=0$ or $y=1$?
* Looking at the grid lines: The x-axis is $y=0$. The curve in H at $x=180$ seems to touch the line $y=1$? No, if it touched $y=1$, the amplitude would be $0.5$?
* Let's look really closely at H.
* Start $(0,2)$. Point $(90,1)$. Point $(180,2)$? Point $(270,3)$. Point $(360,2)$.
* This pattern $2,1,2,3,2$ corresponds to $y = -\cos(x) + 2$?
* $-\cos(0)+2 = 1$. No.
* $y = \sin(x-90) + 2$?
* Let's re-read the options. Maybe I missed one.
* Options:
1. $y = -\cos(x)$ -> Graph G (Starts -1, max 1, min -1). Correct.
2. $y = -\tan(x)$
3. $y = -\sin(x)$ -> Graph F (Starts 0, goes down). Correct.
4. $y = \sin(x) - 3$ -> Graph B (Starts -3, max -2, min -4). Correct.
5. $y = \tan(x - 90)$
6. $y = \cos(x)$ -> Graph C (Starts 1, min -1). Correct.
7. $y = \tan(x)$
8. $y = \tan(x) + 2$
9. $y = \cos(x) - 2$ -> Graph E (Starts -1, min -3, max -1). Correct.
10. $y = \sin(x)$ -> Graph A (Starts 0, max 1, min -1). Correct.
11. $y = \sin(x) + 2$ -> Graph D (Starts 2, max 3, min 1). Correct.
12. $y = \cos(x) + 1$

* So, Graph H MUST be $y = \cos(x) + 1$. Let's force-fit it.
* Equation: $y = \cos(x) + 1$.
* Points: $(0,2), (90,1), (180,0), (270,1), (360,2)$.
* Does Graph H pass through $(180,0)$? Looking at the image provided... In box H, at $x=180$, the red line is distinctly ABOVE the x-axis. It looks like it is at $y=1$ or $y=2$.
* However, let's look at Graph C again. Graph C is $y=\cos(x)$. Range $[-1, 1]$.
* Let's look at Graph H again. Range appears to be $[1, 3]$? If range is $[1,3]$, midline is $2$, amplitude is $1$.
* If midline is $2$ and it starts at $2$ going down, it is $-\sin(x) + 2$. Not an option.
* If it starts at $2$ going down, could it be $\cos(x+90) + 2$? Same thing.
* Is there an error in my reading of Graph H?
* Let's look at Graph L. Tangent.
* Let's look at Graph K. Tangent.
* Let's look at Graph J. Tangent.
* Let's look at Graph I. Tangent.

* Let's re-evaluate Graph H vs the last remaining cosine/sine option: $y = \cos(x) + 1$.
* Perhaps the graph labeled H is actually drawn with a phase shift or I am misinterpreting the grid.
* Let's look at the other graphs to ensure no other mistakes.
* Graph A: $y=\sin(x)$. OK.
* Graph B: $y=\sin(x)-3$. OK.
* Graph C: $y=\cos(x)$. OK.
* Graph D: $y=\sin(x)+2$. OK.
* Graph E: $y=\cos(x)-2$. OK.
* Graph F: $y=-\sin(x)$. OK.
* Graph G: $y=-\cos(x)$. OK.
* This leaves Graph H for $y=\cos(x)+1$. Even if the drawing looks slightly off (the minimum at 180 looks higher than 0), it is the only remaining trigonometric function that fits the "Cosine" family with a vertical shift. Wait, look at Graph H again. Does it cross the x-axis? No. The lowest point is clearly positive. $y=\cos(x)+1$ has a minimum of $0$. It touches the x-axis. In the diagram for H, the curve at $x=180$ is visibly separated from the x-axis.
* Is it possible Graph H is $y = |\cos(x)| + 1$? No absolute value in options.
* Is it possible I swapped C and H?
* Graph C: Starts at $1$. Min at $180$ is $-1$. Crosses axis. Definitely $\cos(x)$.
* Graph H: Starts at $2$. Min at $180$ is... well, if the option is $\cos(x)+1$, it *should* be $0$. Maybe the drawing is just thick/imprecise. Or maybe I am blind. Let's assume H is $y=\cos(x)+1$ by elimination.

2. Analyze the Tangent Graphs (I-L)
Tangent graphs have vertical asymptotes and repeat every $180^\circ$. They pass through the midline at regular intervals.

* Graph I: Passes through $(0,0)$. Goes to $+\infty$ as $x \to 90$. Comes from $-\infty$ as $x \to 90$ from right? No.
* Standard $\tan(x)$ passes through $(0,0)$, goes up to $+\infty$ at $90$, comes from $-\infty$ after $90$, passes through $(180,0)$.
* Graph I: At $x=0, y=0$. At $x=45, y=1$. At $x=90$, asymptote. At $x=135, y=-1$? No, it looks like it passes through $(180,0)$.
* This is the standard tangent graph.
* So, Graph I is $y = \tan(x)$.

* Graph J: Passes through $(0,2)$. Asymptote at $90$?
* Let's check the points. At $x=0, y=2$. At $x=180, y=2$. At $x=360, y=2$.
* This is $\tan(x)$ shifted up by $2$.
* So, Graph J is $y = \tan(x) + 2$.

* Graph K: Passes through $(90,0)$? No, at $x=90$ there is an asymptote?
* Let's look at the intercepts. It crosses the x-axis at $x=0$? No, at $x=0$ it is at $y=-1$? No.
* Let's look at the asymptotes. Vertical lines at $x=0$? No, the curve exists at $x=0$.
* Let's look at Graph K carefully.
* It crosses the x-axis at $x=90$? No, it looks like it crosses at $x=90$.
* It has an asymptote at $x=0$? No.
* It has an asymptote at $x=180$? No.
* Let's look at the shape. It goes from bottom-left to top-right. Positive slope.
* Where are the asymptotes? Looks like $x=0$ is NOT an asymptote. $x=90$ is NOT an asymptote.
* Wait, look at Graph K again. The red line crosses the x-axis at $x=90$ and $x=270$.
* Standard $\tan(x)$ crosses at $0, 180, 360$.
* $\tan(x-90)$ shifts the graph to the right by $90$. So zeros move to $90, 270$. Asymptotes move to $90+90=180$? No. Asymptotes of $\tan(x)$ are at $90, 270$. Shift right by $90 \rightarrow 180, 360$.
* Let's check Graph K for asymptotes. There is a break between the curve ending near $x=180$ and starting again?
* Actually, let's look at Graph L first.

* Graph L: Has asymptotes at $x=90$ and $x=270$.
* The curve goes from Top-Left to Bottom-Right. Negative slope.
* This is a reflected tangent graph. $y = -\tan(x)$.
* Let's verify. $-\tan(45) = -1$. Graph L at $x=45$ is at $y=-1$? Yes. $-\tan(135) = -(-1) = 1$. Graph L at $x=135$ is at $y=1$? Yes.
* So, Graph L is $y = -\tan(x)$.

* Now back to Graph K.
* Remaining tangent options: $y = \tan(x - 90)$.
* Let's analyze $y = \tan(x - 90)$.
* Zeros: $x-90 = 0 \Rightarrow x=90$. $x-90=180 \Rightarrow x=270$.
* Asymptotes: $x-90 = 90 \Rightarrow x=180$. $x-90 = -90 \Rightarrow x=0$.
* So, asymptotes at $x=0$ and $x=180$.
* Let's look at Graph K.
* Does it have an asymptote at $x=0$? The graph starts from the bottom left corner. As $x \to 0^+$, $y \to -\infty$? Yes, the line goes straight down at the y-axis.
* Does it have an asymptote at $x=180$? Yes, the line goes straight down/up around $180$.
* Does it cross zero at $90$? Yes.
* So, Graph K is $y = \tan(x - 90)$.

Summary of Matches:

* A: $y = \sin(x)$
* B: $y = \sin(x) - 3$
* C: $y = \cos(x)$
* D: $y = \sin(x) + 2$
* E: $y = \cos(x) - 2$
* F: $y = -\sin(x)$
* G: $y = -\cos(x)$
* H: $y = \cos(x) + 1$ (By elimination, despite visual ambiguity on the minimum)
* I: $y = \tan(x)$
* J: $y = \tan(x) + 2$
* K: $y = \tan(x - 90)$
* L: $y = -\tan(x)$

Let's double check H.
If H is $y = \cos(x) + 1$, min is 0.
Look at Graph H again. The curve at $x=180$ is the local minimum. The grid line below it is the x-axis ($y=0$). The curve seems to rest *on* the grid line $y=1$?
Wait. Look at Graph C. Min is $-1$.
Look at Graph G. Min is $-1$. Start is $-1$. Max is $1$.
Look at Graph H. Start is $2$. If it were $\cos(x)+1$, min is $0$.
Is there any other option?
We used:
$\sin(x)$ (A)
$\sin(x)-3$ (B)
$\cos(x)$ (C)
$\sin(x)+2$ (D)
$\cos(x)-2$ (E)
$-\sin(x)$ (F)
$-\cos(x)$ (G)
$\tan(x)$ (I)
$\tan(x)+2$ (J)
$\tan(x-90)$ (K)
$-\tan(x)$ (L)

The only unused equation is $y = \cos(x) + 1$.
Therefore, H must be $y = \cos(x) + 1$. The drawing might just be slightly inaccurate regarding the depth of the trough, or the line thickness makes it look higher than it is. Mathematically, it is the only fit.

Final Answer:
A: y = sin(x)
B: y = sin(x) - 3
C: y = cos(x)
D: y = sin(x) + 2
E: y = cos(x) - 2
F: y = -sin(x)
G: y = -cos(x)
H: y = cos(x) + 1
I: y = tan(x)
J: y = tan(x) + 2
K: y = tan(x - 90)
L: y = -tan(x)