# Drawing sine and cosine waves

I like mathematics and pretty much every mathematical subject, but if there is one thing I thoroughly dislike, it is drawing (functions, waves, diagrams, etc.) We have this important trig test coming up and I need to master the drawing of sine and cosine waves. Can you guys give me an action plan of how to draw (co)sine waves? Like what to do first, second and so on. I missed a month of school because of my pneumonia so I really need all the help I can get (outside of school).

## 4 Answers

Once you're comfortable with the information in the earlier posts, you can graph more complicated functions of the form $f(x)=a\sin{(b(x-c))}+d$. The parameter $d$ quantifies vertical translation; $a$ is a vertical dilation (amplitude of the wave). The parameter $b$ corresponds to horizontal dilation, whereas $c$ corresponds to horizontal translation.

For example, to graph $f(x) = 3\sin{(2(x-\frac{\pi}{2}))}-1$, you can translate the graph of $y=\sin(x)$ by $\pi/2$ units to the right, then halve its period (double the number of peaks & troughs in an interval of length $2\pi$), stretch that vertically by multiplying every y-coordinate by three, and then translating the whole graph down one unit. I'm too new a user to be able to post a picture, but if you want visualization, you can visit Wolfram Alpha (link)

Remember that the shapes(amplitude and wavelength) of both waves are essentially the same; however, there is a difference of $\pi/2$radians.

The following picture shows a sine wave: ... and the cosine wave: Notice how at $0 \text{ radians}$, the sine wave is at the 'bottom' of the graph(zero) and the cosine wave is at the 'extreme' of the graph (1).

Therefore, they are related... we have just shifted the sine graph by $\pi/2$ radians to the right.(or to the left, we may use interchangeably!)

Recall the trigonometric identity $\cos(90^{\circ} - x) = \sin (x)$. This trig identity is based on the above fact!

Conclusion: to memorize both waves, just memorize one and shift it by $\pi/2$ or $90^{\circ}$.

Edit— for visualization help, here is the plotting of both on the same graph: It's easy to recognize that purple is $\sin(x)$ and blue is $\cos(x)$.

One important point which hasn't yet been mentioned is that the sine has slope 1 at the origin, that is, it has the tangent $y=x$, which is the diagonal (the plots by Parth Kohli use different scales for the $x$ and $y$ axis, therefore you don't easily see that there). The same happens of course also at all multiples of $2\pi$. Moreover, at odd multiples of $\pi$ you get the reverse slope.

Add to that that the value at $\pi/2$ is $1$ and at $3\pi/2$ is $-1$, and that those are the extremal values, and you'll find that this already characterizes the curve quite well: If you have $a\sin(bx)$, the slopes are $ab$ instead of $1$, the extremes have value $\pm a$ and the zeros are at multiples of $\pi/b$ instead (the extrema are of course always exactly in between two zeros).

If you get the (x,y) and slope correct at the zeros of the first few derivatives then this is enough.