Graphs of sine and cosine
Theory
The graphs of \(y=\sin x\) and \(y=\cos x\) are smooth, repeating waves, each with amplitude \(1\) and period \(2\pi\), oscillating between \(-1\) and \(1\). Reading the key points — the maxima, minima and zeros — lets you sketch either curve and solve simple equations by eye, and the two graphs are linked by \(\cos x=\sin\left(x+\dfrac{\pi}{2}\right)\). Angles are measured in radians throughout.
The sine curve \(y=\sin x\) is defined for every real \(x\) and takes values in the range \([-1,1]\). Its amplitude — the distance from the centre line to a peak — is \(1\), and its period is \(2\pi\), meaning one complete cycle occurs each time \(x\) increases by \(2\pi\). Starting at the origin, the curve rises to a maximum of \(1\) at \(x=\dfrac{\pi}{2}\), returns to \(0\) at \(x=\pi\), falls to a minimum of \(-1\) at \(x=\dfrac{3\pi}{2}\), and returns to \(0\) at \(x=2\pi\).
The cosine curve \(y=\cos x\) has the same amplitude \(1\) and period \(2\pi\), but it begins at its maximum: it passes through \((0,1)\), falls to \(0\) at \(x=\dfrac{\pi}{2}\), reaches a minimum of \(-1\) at \(x=\pi\), returns to \(0\) at \(x=\dfrac{3\pi}{2}\), and climbs back to \(1\) at \(x=2\pi\). Its zeros are therefore the odd multiples of \(\dfrac{\pi}{2}\), while the zeros of \(\sin x\) are the whole multiples of \(\pi\).
Both curves are symmetric. Sine is an odd function, so \(\sin(-x)=-\sin x\) and its graph has point symmetry about the origin; cosine is an even function, so \(\cos(-x)=\cos x\) and its graph is symmetric about the \(y\)-axis. The two are a quarter-turn apart: shifting the sine graph \(\dfrac{\pi}{2}\) to the left gives the cosine graph, because \(\cos x=\sin\left(x+\dfrac{\pi}{2}\right)\).
Both curves share the same amplitude, period and range:
The key values of \(\sin x\) over one period:
The key values of \(\cos x\) over one period:
Symmetry (sine odd, cosine even) and the shift that links the two graphs:
How to sketch \(y=\sin x\) or \(y=\cos x\) and read from it
- Set up the axes. Draw the \(x\)-axis over one period \(0\le x\le 2\pi\) and mark the quarter-points \(\dfrac{\pi}{2},\ \pi,\ \dfrac{3\pi}{2},\ 2\pi\); draw the amplitude lines \(y=1\) and \(y=-1\).
- Plot the five key points. For \(\sin x\) use \((0,0),\left(\dfrac{\pi}{2},1\right),(\pi,0),\left(\dfrac{3\pi}{2},-1\right),(2\pi,0)\); for \(\cos x\) start at the maximum \((0,1)\) and step down through \(\left(\dfrac{\pi}{2},0\right),(\pi,-1),\left(\dfrac{3\pi}{2},0\right),(2\pi,1)\).
- Join with a smooth wave. Curve through the points — rounded at the peaks and troughs, never a sharp corner — and extend by repeating every \(2\pi\) if more of the graph is needed.
- Read values or solve. To evaluate, read the height at the required \(x\); to solve an equation, draw the horizontal line at that value and read the \(x\)-coordinates of every intersection inside the interval.
| amplitude | \(=\) | \(1\) |
| period | \(=\) | \(2\pi\) |
| range | \(=\) | \([-1,1]\) |
| maximum | \(=\) | \(\left(\dfrac{\pi}{2},\,1\right)\) |
| minimum | \(=\) | \(\left(\dfrac{3\pi}{2},\,-1\right)\) |
| \(\cos x\) | \(=\) | \(0\) |
| \(x\) | \(=\) | \(\dfrac{\pi}{2},\ \dfrac{3\pi}{2}\) |
| maxima | \(=\) | \((0,1)\text{ and }(2\pi,1)\) |
| minimum | \(=\) | \((\pi,-1)\) |
| \(x\) | \(=\) | \(0,\ \pi,\ 2\pi\) |
| \(x\) | \(=\) | \(\dfrac{3\pi}{2}\) |
| \(\cos x\) | \(=\) | \(\sin\!\left(x+\dfrac{\pi}{2}\right)\) |
| \(\sin x\) | \(=\) | \(\cos x\) |
| \(\tan x\) | \(=\) | \(1\) |
| \(x\) | \(=\) | \(\dfrac{\pi}{4},\ \dfrac{5\pi}{4}\) |
Common pitfalls
Frequently asked questions
What does the graph of y = sin x look like?
It is a smooth wave. Starting from the origin it rises to a maximum of \(1\) at \(x=\dfrac{\pi}{2}\), returns to \(0\) at \(x=\pi\), falls to a minimum of \(-1\) at \(x=\dfrac{3\pi}{2}\), and returns to \(0\) at \(x=2\pi\). This one cycle then repeats every \(2\pi\) in both directions.
What are the amplitude and period of y = sin x and y = cos x?
Both have amplitude \(1\), so they oscillate between \(y=-1\) and \(y=1\), and both have period \(2\pi\), meaning one full cycle is completed as \(x\) increases by \(2\pi\). The amplitude is the height from the centre line to a peak.
Where are the maximum, minimum and zeros of y = cos x?
On the interval from \(0\) to \(2\pi\), \(y=\cos x\) has a maximum of \(1\) at \(x=0\) and \(x=2\pi\), a minimum of \(-1\) at \(x=\pi\), and zeros at \(x=\dfrac{\pi}{2}\) and \(x=\dfrac{3\pi}{2}\). The cosine graph therefore begins at its peak.
How are the graphs of sine and cosine related?
The cosine graph is the sine graph translated \(\dfrac{\pi}{2}\) to the left, because \(\cos x=\sin\left(x+\dfrac{\pi}{2}\right)\). Equivalently the sine graph is the cosine graph translated \(\dfrac{\pi}{2}\) to the right, since \(\sin x=\cos\left(x-\dfrac{\pi}{2}\right)\).
How do you solve an equation such as sin x = 1 or cos x = 0 from the graph?
Draw the horizontal line at that value and read off the \(x\)-coordinates where it meets the curve inside the required interval. For \(\sin x=1\) on \(0\) to \(2\pi\) the only solution is \(x=\dfrac{\pi}{2}\), and for \(\cos x=0\) the solutions are \(x=\dfrac{\pi}{2}\) and \(x=\dfrac{3\pi}{2}\).
Is sine an odd function and cosine an even function?
Yes. Sine is odd, so \(\sin(-x)=-\sin x\) and its graph has point symmetry about the origin. Cosine is even, so \(\cos(-x)=\cos x\) and its graph is symmetric about the \(y\)-axis.