Trigonometric Functions and Signs in Quadrants

Introduction

Trigonometric functions extend the concept of ratios in a right-angled triangle to any angle. This is achieved by using the unit circle, which allows us to define these functions for angles beyond the acute angles found in a triangle.

The Unit Circle Definition

A unit circle is a circle with a radius of 1, centered at the origin of the Cartesian plane. For any angle $\theta$ measured counter-clockwise from the positive x-axis, the point P(x, y) where its terminal side intersects the unit circle gives us the primary trigonometric values.

• Cosine ($\cos\theta$) is defined as the x-coordinate of the point P.

• Sine ($\sin\theta$) is defined as the y-coordinate of the point P.

The other four trigonometric functions are defined in terms of sine and cosine:

• $$\tan\theta = \frac{y}{x} = \frac{\sin\theta}{\cos\theta} \quad (x \ne 0)$$

• $$\sec\theta = \frac{1}{x} = \frac{1}{\cos\theta} \quad (x \ne 0)$$

• $$\csc\theta = \frac{1}{y} = \frac{1}{\sin\theta} \quad (y \ne 0)$$

• $$\cot\theta = \frac{x}{y} = \frac{\cos\theta}{\sin\theta} \quad (y \ne 0)$$

Signs of Trigonometric Functions in Quadrants

The signs of the trigonometric functions in each of the four quadrants are determined by the signs of the x and y coordinates. A useful mnemonic to remember this is the ASTC rule ("All Students Take Calculus") or the CAST rule, starting from Quadrant IV.

• Quadrant I ($0^\circ$ to $90^\circ$): All functions are positive (x and y are positive).

• Quadrant II ($90^\circ$ to $180^\circ$): Sine and its reciprocal, Cosecant, are positive (y is positive, x is negative).

• Quadrant III ($180^\circ$ to $270^\circ$): Tangent and its reciprocal, Cotangent, are positive (x and y are negative).

• Quadrant IV ($270^\circ$ to $360^\circ$): Cosine and its reciprocal, Secant, are positive (x is positive, y is negative).

  

Domain and Range of Trigonometric Functions

Function	Domain (Input values of $\theta$)	Range (Output values)
$\sin\theta$	All real numbers ($\mathbb{R}$)	$[-1, 1]$
$\cos\theta$	All real numbers ($\mathbb{R}$)	$[-1, 1]$
$\tan\theta$	$\mathbb{R} - \{(2n+1)\frac{\pi}{2}, n \in \mathbb{Z}\}$ (all reals except odd multiples of $\pi/2$)	All real numbers ($\mathbb{R}$)
$\csc\theta$	$\mathbb{R} - \{n\pi, n \in \mathbb{Z}\}$ (all reals except integer multiples of $\pi$)	$(-\infty, -1] \cup [1, \infty)$
$\sec\theta$	$\mathbb{R} - \{(2n+1)\frac{\pi}{2}, n \in \mathbb{Z}\}$ (all reals except odd multiples of $\pi/2$)	$(-\infty, -1] \cup [1, \infty)$
$\cot\theta$	$\mathbb{R} - \{n\pi, n \in \mathbb{Z}\}$ (all reals except integer multiples of $\pi$)	All real numbers ($\mathbb{R}$)

Example 1: Finding Values from a Point

Question: If the terminal side of an angle $\theta$ passes through the point (-4, 3), find the values of all six trigonometric functions.

Solution:

This point is not on the unit circle. We have $x=-4$ and $y=3$. First, we find the distance 'r' from the origin to this point, which acts as the hypotenuse.

$r = \sqrt{x^2+y^2} = \sqrt{(-4)^2+3^2} = \sqrt{16+9} = \sqrt{25}=5$.

The definitions are then generalized as follows:

• $\sin\theta = y/r = \mathbf{3/5}$

• $\cos\theta = x/r = \mathbf{-4/5}$

• $\tan\theta = y/x = 3/(-4) = \mathbf{-3/4}$

• $\csc\theta = 1/\sin\theta = r/y = \mathbf{5/3}$

• $\sec\theta = 1/\cos\theta = r/x = 5/(-4) = \mathbf{-5/4}$

• $\cot\theta = 1/\tan\theta = x/y = -4/3 = \mathbf{-4/3}$

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Example 2: Finding Values from One Ratio and Quadrant

Question: If $\cos x = -3/5$ and x lies in the third quadrant, find the values of the other five trigonometric functions.

Solution:

Step 1: Use the Pythagorean identity $\sin^2x + \cos^2x = 1$ to find $\sin x$.

$\sin^2x = 1 - (-3/5)^2 = 1 - 9/25 = 16/25$. So, $\sin x = \pm 4/5$.

Since x is in the third quadrant, $\sin x$ is negative. Thus, $\mathbf{\sin x = -4/5}$.

Step 2: Use the found values of sin and cos to find the rest.

• $\tan x = \frac{\sin x}{\cos x} = \frac{-4/5}{-3/5} = \mathbf{4/3}$.

• $\csc x = 1/\sin x = \mathbf{-5/4}$.

• $\sec x = 1/\cos x = \mathbf{-5/3}$.

• $\cot x = 1/\tan x = \mathbf{3/4}$.

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Example 3: Evaluating at Large Angles

Question: Find the value of $\sin(765^\circ)$.

Solution:

Trigonometric functions are periodic. The period of sine is $360^\circ$. We can add or subtract any multiple of $360^\circ$ without changing the value. We find how many full revolutions are in $765^\circ$.

$765 \div 360 = 2$ with a remainder. $765^\circ = 2 \times 360^\circ + 45^\circ$.

Therefore, $\sin(765^\circ) = \sin(45^\circ) = \mathbf{\frac{1}{\sqrt{2}}}$.

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Example 4: Evaluating with Negative Angles in Radians

Question: Find the value of $\cot(-15\pi/4)$.

Solution:

Step 1: Use the identity $\cot(-\theta) = -\cot(\theta)$. So we need to find $-\cot(15\pi/4)$.

Step 2: Simplify the angle. $15\pi/4 = (16\pi - \pi)/4 = 4\pi - \pi/4$.

So, we need to find $-\cot(4\pi - \pi/4)$. The period of cotangent is $\pi$, so we can ignore the $4\pi$.

$-\cot(-\pi/4) = -(-\cot(\pi/4)) = \cot(\pi/4) = \mathbf{1}$.

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Example 5: Finding an Angle from Signs

Question: If $\sin x = 1/2$ and $\cos x = -\sqrt{3}/2$, find the principal value of x.

Solution:

We are given that $\sin x$ is positive and $\cos x$ is negative. According to the ASTC rule, this occurs only in the second quadrant. The reference angle (the acute angle with the x-axis) for which $\sin x = 1/2$ is $30^\circ$ or $\pi/6$. The corresponding angle in the second quadrant is $180^\circ - 30^\circ = 150^\circ$, or in radians, $\pi - \pi/6 = 5\pi/6$. The principal value is $5\pi/6$.

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Example 6: Determining the Sign of an Expression

Question: Determine the sign of the expression $\sin(2) \cdot \cos(3) \cdot \tan(4)$, where the angles are in radians.

Solution:

We need to determine the quadrant for each angle. We use the approximations $\pi \approx 3.14$ and $\pi/2 \approx 1.57$.

• 2 radians: Since $1.57 < 2 < 3.14$ (i.e., $\pi/2 < 2 < \pi$), this angle is in Quadrant II. In Q2, $\sin(2)$ is positive.

• 3 radians: Since $1.57 < 3 < 3.14$ (i.e., $\pi/2 < 3 < \pi$), this angle is in Quadrant II. In Q2, $\cos(3)$ is negative.

• 4 radians: Since $3.14 < 4 < 3\pi/2 \approx 4.71$ (i.e., $\pi < 4 < 3\pi/2$), this angle is in Quadrant III. In Q3, $\tan(4)$ is positive.

The sign of the entire expression is $(+) \times (-) \times (+) = (-)$. The expression is negative.

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Example 7: Range of a Trigonometric Function

Question: What is the range of the function $f(x) = 5\cos(2x)-3$?

Solution:

Step 1: Start with the range of the basic cosine function: $-1 \le \cos(2x) \le 1$.

Step 2: Multiply by 5 (the amplitude): $-5 \le 5\cos(2x) \le 5$.

Step 3: Subtract 3 (the vertical shift): $-5-3 \le 5\cos(2x)-3 \le 5-3$.

This simplifies to $-8 \le f(x) \le 2$. The range is $[-8, 2]$.

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Example 8: Evaluating a Complex Expression

Question: Find the value of $\sec(270^\circ - \theta) \sec(90^\circ - \theta) - \tan(270^\circ - \theta) \tan(90^\circ + \theta)$.

Solution:

We simplify each term using allied angle rules (ASTC and co-function identities):

• $\sec(270^\circ - \theta) = -\csc\theta$ (Q3, secant is negative, changes to cosecant)

• $\sec(90^\circ - \theta) = \csc\theta$ (Q1, secant is positive, changes to cosecant)

• $\tan(270^\circ - \theta) = \cot\theta$ (Q3, tangent is positive, changes to cotangent)

• $\tan(90^\circ + \theta) = -\cot\theta$ (Q2, tangent is negative, changes to cotangent)

The expression becomes $(-\csc\theta)(\csc\theta) - (\cot\theta)(-\cot\theta) = -\csc^2\theta + \cot^2\theta = -(\csc^2\theta - \cot^2\theta)$.

Using the Pythagorean identity $1+\cot^2\theta=\csc^2\theta$, we know $\csc^2\theta - \cot^2\theta = 1$. The final value is -1.