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Tangent of Acute Angles


Alignments to Content Standards: G-SRT.C.6

Task

  1. Below is a picture of a right triangle:


    In terms of the picture, what are $\sin{\angle P}$, $\cos{\angle P}$, and $\tan{\angle P}$? Do these values depend on the size of the triangle?

  2. Complete the following table, rounding off each answer to the nearest hundredth if using a calculator. Draw a picture showing the meaning of  $\sin{x}$, $\cos{x}$, and $\tan{x}$ for an acute angle $x$.

     

    Angle (degrees) $\cos{x}$ $\sin{x}$ $\tan{x}$
    $0$      
    $15$      
    $30$      
    $45$      
    $60$      
    $75$      
    $90$      

     

  3. What value do you find in the bottom right corner of the table for $\tan{90}$? Why?
  4. What patterns do you notice in the third column with the values of $\tan{x}$? Do you think that these patterns will hold true for all acute angles $x$? Explain.
  5. Based on the table, what values do you think the function $\tan{x}$ takes when $0 \lt x \lt 90$? Explain.

IM Commentary

The purpose of this task is to focus on studying values of $\tan{x}$ for special angles and conjecturing from these values how the function $\tan{x}$ varies when $0 \leq x \lt 90$. This task complements https://www.illustrativemathematics.org/illustrations/1868 which studies the same table for $\sin{x}$ and $\cos{x}$ but looks at some other expressions in terms of $\sin{x}$ and $\cos{x}$ rather than $\tan{x}$. The teacher might also wish to add another column for the cotangent function as students should be able to conjecture and show that $\tan{x} = \cot{(90-x)}$ for acute angles $x$. Exact values for $\sin{x}$ and $\cos{x}$ can be found in the commentary to the aforementioned task and these could be used to obtain exact values of $\tan{x}$ and $\cot{x}$. Our practice in the solution to this task is to put in exact values for benchmark angles and approximate decimal values for the non benchmark angles.  Exact such values can be found, for example, from half-angle formulas applied to known values (e.g., applying the sine half-angle formula to compute $\sin(15^\circ)$ from $\sin(30^\circ)$).

Solution

  1. The sine of $\angle P$ is the length of the side opposite $P$, $|QR|$, divided by the hypotenuse, $|PR|$: $$\sin{P} = \frac{|QR|}{|PR|}.$$The cosine of $\angle P$ is the length of the side adjacent to $P$, $|PQ|$, divided by the length of the hypotenuse, $|PR|$: $$\cos{P} = \frac{|PQ|}{|PR|}.$$The tangent of $\angle P$ is the length of the side opposite $P$, $|QR|$ divided by the side adjacent to $P$, $|PQ|$: $$\tan{P} = \frac{|QR|}{|PQ|}.$$

    These ratios do not depend on the size of the triangle. If the triangle is scaled by a (positive) factor of $r$ then all three side lengths scale by $r$. The three trigonometric ratios are scaled by a factor of $\frac{r}{r} = 1$ and so they do not depend on the size of the triangle.

  2. Exact values are entered in each row except for 15$^\circ$ and 75$^\circ$:

    Angle (degrees) $\cos{x}$ $\sin{x}$ $\tan{x}$
    $0$ 1 0 0
    $15$ 0.97 0.26 0.27
    $30$ $\frac{\sqrt{3}}{2}$ 0.5 $\frac{1}{\sqrt{3}}$
    $45$ $\frac{\sqrt{2}}{2}$ $\frac{\sqrt{2}}{2}$ 1
    $60$ 0.5 $\frac{\sqrt{3}}{2}$ $\sqrt{3}$
    $75$ 0.26 0.97 3.7
    $90$ 0 1 --

    Note that there is no entry for $\tan{90}$ because $\frac{1}{0}$ is not defined.

  3. We have $\tan{x} = \frac{\sin{x}}{\cos{x}}$ and this function is not defined when the denominator, $\cos{x}$, is zero. This happens, on the unit circle, for $x = 90$ and $x = 270$. For all other angles $0 \leq x \lt 360$, $\tan{x}$ has a well defined value.  
  4. There are a few patterns. First, the values of $\tan{x}$ for $0 \leq x \lt 90$ appear to be non-negative and the value increases as $x$ increases. Further experimentation will show that this pattern continues. For example, $\tan{80} \approx 5.7$, $\tan{85} \approx 11$, and $\tan{89} \approx 57$. This makes sense since in a right triangle with an 89 degree angle and a 1 degree angle, the side opposite the 89 degree angle is much larger than the side adjacent and this disparity grows as the 89 angle grows closer and closer to 90 degrees. 

    Another pattern is perhaps harder to see. We can see that $\tan{60} = \frac{1}{\tan{30}}$ and, if we had exact values, we also have $\tan{75} = \frac{1}{\tan{15}}$. In general for a (non-zero) acute angle $x$, we have $\tan{x} = \frac{1}{\tan{(90-x)}}$. This comes from the identities $\sin{x} = \cos{(90-x)}$ and $\cos{x} = \sin{(90-x)}$. To see why, note that

    $$\begin{align} \tan{x} &= \frac{\sin{x}}{\cos{x}} \\ &= \frac{\cos{(90-x)}}{\sin{(90-x)}} \\ &= \frac{1}{\tan{(90-x)}} \end{align}$$
  5. We can see that $\tan{0} = 0$ and we have also seen that for $0 \leq x < 90$, $\tan{x}$ grows as $x$ increases. There is no bound to how big $\tan{x}$ can be because when we write it as $\frac{\sin{x}}{\cos{x}}$ we can see that for acute angles just a little bit less than 90 degrees, the numerator of this fraction, $\sin{x}$, is close to 1 while the denominator, $\cos{x}$ is very close to zero. So the range of $\tan{x}$ for $0 \lt x \lt 90$ is all non-negative real numbers.