Chapter 5.1, Problem 36E

To determine

To find: the reference number for each value of angle t.

Answer to Problem 36E

  $\left(\text{a}\right)$$\overline{\text{t}}\text{=}\frac{\text{π}}{6}$ , $\left(\text{b}\right)$$\overline{\text{t}}\text{=}\frac{\text{π}}{6}$ , $\left(\text{c}\right)$$\overline{\text{t}}\text{=}\frac{\text{π}}{\text{3}}$ and $\left(\text{d}\right)$$\overline{\text{t}}\text{=}\frac{\text{π}}{4}$ .

Explanation of Solution

Given:

  $\left(\text{a}\right)$$\text{t=}\frac{\text{5π}}{6}$ . $\left(\text{b}\right)$$\text{t=}$$\frac{\text{7π}}{6}$ .

  $\left(\text{c}\right)$$\text{t=}\frac{\text{11π}}{3}$ . $\left(\text{d}\right)$$\text{t=-}\frac{\text{7π}}{4}$ .

Concept used:

For a unit circle:

  $\text{r=1}$ .

  $\text{x=cosθ}$ .

  $\text{y=sinθ}$ .

The reference angle should be less than $\frac{\text{π}}{\text{2}}$ .

The angle is always considered or calculated form x-axis.

Due to periodic nature of sine and cosine function the value of the angle repeat again and again therefore the higher value of angle can be interprets in-term of reference angle which is less than ${\text{90}}^0$ .

Calculation:

  $\left(\text{a}\right)$$\text{t=}\frac{\text{5π}}{6}$

Which can be written as:

  $\text{t=}\frac{\text{5π}}{6}\text{=π-}\frac{\text{π}}{6}\text{=}\frac{\text{π}}{6}$ .

Since during horizontal shift trigonometric ratio did not change and its reference angle will be generated.

  $\text{T}\left(\text{π-θ}\right)\text{=T}\left(\text{θ}\right)$ .

Let T be any trigonometric ratio and irrespective of trigonometric all give same reference angle.

Therefore

  $\left(\text{a}\right)$$\overline{\text{t}}\text{=}\frac{\text{π}}{6}$ .

Likewise,

  $\left(\text{b}\right)$

  $\frac{\text{7π}}{6}$ .

Which can be written as:

  $\text{t=}\frac{\text{7π}}{6}\text{=π+}\frac{\text{π}}{6}\text{=}\frac{\text{π}}{6}$ .

Since during horizontal shift trigonometric ratio did not change and its reference angle will be generated.

Therefore,

  $\left(\text{b}\right)$

  $\overline{\text{t}}\text{=}\frac{\text{π}}{6}$ .

Likewise,

  $\left(\text{c}\right)$$\text{t=}\frac{\text{11π}}{\text{3}}$ .

Which can be written as:

  $\text{t=}\frac{\text{11π}}{\text{3}}\text{=4π-}\frac{\text{π}}{\text{3}}\text{=}\frac{\text{π}}{\text{3}}$ .

Therefore,

  $\left(\text{c}\right)$$\overline{\text{t}}\text{=}\frac{\text{π}}{\text{3}}$ .

Since during horizontal shift trigonometric ratio did not change and its reference angle will be generated.

  $\left(\text{d}\right)$$\text{t=-}\frac{\text{7π}}{4}$ .

  $\text{t=-}\frac{\text{7π}}{4}\text{=-2π+}\frac{\text{π}}{4}\text{=}\frac{\text{π}}{\text{4}}$ .

Since the angle lies in 1^st quadrant the angle its reference angle

Therefore,

  $\left(\text{d}\right)$$\overline{\text{t}}\text{=}\frac{\text{π}}{4}$ .

Hence, $\left(\text{a}\right)$$\overline{\text{t}}\text{=}\frac{\text{π}}{6}$ , $\left(\text{b}\right)$$\overline{\text{t}}\text{=}\frac{\text{π}}{6}$ , $\left(\text{c}\right)$$\overline{\text{t}}\text{=}\frac{\text{π}}{\text{3}}$ and $\left(\text{d}\right)$$\overline{\text{t}}\text{=}\frac{\text{π}}{4}$ .