# 1.5 Limits Of Trig Functionsap Calculus

1-4 Trigonometry Appendix D 1-5. Function Concept and Representations; Domain and Range: 1.1. Essential Elementary Functions; New Functions from Old 1.2-1.3. Exponentials, Inverse Functions and Logarithms. 1.5, 1.6: 1-9 Limit Concept; Limit Calculation Laws 2.1-2.3 1-10 Continuity 2.5. Math AP®︎/College Calculus AB Limits and continuity Determining limits using algebraic properties of limits: direct substitution Limits of trigonometric functions AP.CALC: LIM‑1 (EU), LIM‑1.D (LO), LIM‑1.D.1 (EK).

### The Six Basic Trigonometric Functions

Trigonometric functions allow us to use angle measures, in radians or degrees, to find the coordinates of a point on any circle—not only on a unit circle—or to find an angle given a point on a circle. They also define the relationship among the sides and angles of a triangle.

To define the trigonometric functions, first consider the unit circle centered at the origin and a point (P=(x,y)) on the unit circle. Let (θ) be an angle with an initial side that lies along the positive (x)-axis and with a terminal side that is the line segment (OP). An angle in this position is said to be in standard position (Figure (PageIndex{2})). We can then define the values of the six trigonometric functions for (θ) in terms of the coordinates (x) and (y).

Definition: T**rigonometric functions**

Let (P=(x,y)) be a point on the unit circle centered at the origin (O). Let (θ) be an angle with an initial side along the positive (x)-axis and a terminal side given by the line segment (OP). The trigonometric functions are then defined as

(sin θ=y) | (cscθ=dfrac{1}{y}) |

(cos θ=x) | (secθ=dfrac{1}{x}) |

(tan θ=dfrac{y}{x}) | (cot θ=dfrac{x}{y}) |

If (x=0, secθ) and (tan θ) are undefined. If (y=0), then (cot θ) and (cscθ) are undefined.

We can see that for a point (P=(x,y)) on a circle of radius (r) with a corresponding angle (θ), the coordinates (x) and (y) satisfy

[begin{align} cos θ &=dfrac{x}{r} x&=rcos θ end{align}]

and

## 1.5 Limits Of Trig Functionsap Calculus Identities

[begin{align} sin θ&=dfrac{y}{r} y&=rsin θ. end{align}]

* Figure (PageIndex{3.1}): *Diagram demonstrating trigonometric functions in the unit circle.,

*).*

The values of the other trigonometric functions can be expressed in terms of (x,y), and (r) (Figure (PageIndex{3})).

**Figure (PageIndex{3.2})**: For a point (P=(x,y)) on a circle of radius (r), the coordinates (x) and y satisfy (x=rcos θ) and (y=rsin θ).

Table shows the values of sine and cosine at the major angles in the first quadrant. From this table, we can determine the values of sine and cosine at the corresponding angles in the other quadrants. The values of the other trigonometric functions are calculated easily from the values of (sin θ) and (cos θ.)

(θ)θ | (sin θ) | (cos θ) |
---|---|---|

0 | 0 | 1 |

(dfrac{π}{6}) | (dfrac{1}{2}) | (dfrac{sqrt{3}}{2}) |

(dfrac{π}{4}) | (dfrac{sqrt{2}}{2}) | (dfrac{sqrt{2}}{2}) |

(dfrac{π}{3}) | (dfrac{sqrt{3}}{2}) | (dfrac{1}{2}) |

(dfrac{π}{2}) | 1 | 0 |

Example (PageIndex{2}): Evaluating Trigonometric Functions

Evaluate each of the following expressions.

- (sin(dfrac{2π}{3}))
- (cos(−dfrac{5π}{6}))
- (tan(dfrac{15π}{4}))

**Solution**:

a) On the unit circle, the angle (θ=dfrac{2π}{3}) corresponds to the point ((−dfrac{1}{2},dfrac{sqrt{3}}{2})). Therefore, (sin)((dfrac{2π}{3}))=(y)=(dfrac{sqrt{3}}{2}).

b) An angle (θ=−dfrac{5π}{6}) corresponds to a revolution in the negative direction, as shown. Therefore, (cos(−dfrac{5π}{6}))=(x)=(−dfrac{sqrt{3}}{2}).

c) An angle (θ)=(dfrac{15π}{4})=(2π)+(dfrac{7π}{4}). Therefore, this angle corresponds to more than one revolution, as shown. Knowing the fact that an angle of (dfrac{7π}{4}) corresponds to the point ((dfrac{sqrt{2}}{2},dfrac{sqrt{2}}{2})), we can conclude that (tan(dfrac{15π}{4})=dfrac{y}{x}=−1).

Exercise (PageIndex{2})

## 1.5 Limits Of Trig Functions Ap Calculus Algebra

Evaluate (cos(3π/4)) and (sin(−π/6)).

## Limits Of Trig Functions Pdf

Look at angles on the unit circle.

**Answer**