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Learning Introductory Physics with Activities

Paul J. Emigh, Rebecka Tumblin, Kathryn Hadley, Danielle Skinner

Section 1.7 Trigonometric Relations

Recall that the sides of a right triangle can be related to the angle using the trigonometric relations in the figure below.
Figure 1.7.1. A right triangle with angle theta and the lengths of the sides labeled as opposite, adjacent and hypotenuse next to a 2D position vector oriented at the origin of an xy-coordinate system.
Consider the sine of angle \(\theta\) and its relationship to the sides of the triangle
\begin{equation*} \sin \theta = \frac{\mathrm{opposite}}{\mathrm{hypotenuse}} = \frac{r_y}{r} \end{equation*}
Suppose you are given the magnitude and direction angle and you would like to find the components of the vector. You can solve the above equation for the \(y\)-component of the position vector \(\vec{r} \) using the above relationship where \(r_y = r\sin\theta\text{.}\) In this case, the \(y\)-component is the magnitude of the vector multiplied by the sine of the angle. 
Now, consider the cosine of angle \(\theta\) and its relationship to the sides of the triangle
\begin{equation*} \cos \theta = \frac{\mathrm{adjacent}}{\mathrm{hypotenuse}} = \frac{r_x}{r} \end{equation*}
To find the \(x\)-component of the position vector \(\vec{r} \) you can use the above relationship where \(r_x = r\cos\theta\text{.}\) 
The third relationship allows you to determine the direction angle \(\theta\) of the position vector using its components.
\begin{equation*} tan \theta = \frac{\mathrm{opposite}}{\mathrm{adjacent}} = \frac{r_y}{r_x} \rightarrow \theta = \tan^{-1}(\frac{r_y}{r_x}) \end{equation*}
Note that vectors are intimately related to right triangle trigonometry. In fact, geometrically the magnitude of a vector is simply the hypotenuse, which can be determined using the familiar Pythagorean theorem.
\begin{equation*} r = |\vec{r}| = \sqrt{r_x^2 + r_y^2} \end{equation*}

Exercises Practice Activities

1.

Find the magnitudes of the vectors \(\vec{v_1}= 3\hat{x}-4\hat{y}\) and \(\vec{v_2}= -12\hat{x}-5\hat{y}\text{.}\) Find their directions as angles relative to an axis of your choice.
Answer 1.
\(|\vec{v_1}|= 5\) at \(-53^o\) from the positive \(x\)-axis.
Answer 2.
\(|\vec{v_2}|= 13\) at \(202.6^o\) from the positive \(x\)-axis.

2.

A vector with length \(12 \mathrm{~m}\) points \(10\) degrees west of south. Sketch this vector and find its \(x\)- and \(y\)-components.

References References

[1]
Going between representations by Dr. Michelle Tomasik from MIT 8.01 Classical Mechanics, Fall 2016, used under Creative Commons BY-NC-SA.
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