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3.1: Fundamental Identities

Created by: CK-12

Introduction

We now enter into the proof portion of trigonometry. Starting with the basic definitions of sine, cosine, and tangent, identities (or fundamental trigonometric equations) emerge. Students will learn how to prove certain identities, using other identities and definitions. Finally, students will be able solve trigonometric equations for theta, also using identities and definitions.

Learning Objectives

• use the fundamental identities to prove other identities.
• apply the fundamental identities to values of $\theta$ and show that they are true.

Quotient Identity

In Chapter 1, the three fundamental trigonometric functions sine, cosine and tangent were introduced. All three functions can be defined in terms of a right triangle or the unit circle.

$\sin \theta & = \frac{opposite}{hypotenuse} = \frac{y}{r} = \frac{y}{1} = y \\\cos \theta & = \frac{adjacent}{hypotenuse} = \frac{x}{r} = \frac{x}{1} = x \\\tan \theta & = \frac{opposite}{adjacent} = \frac{y}{x} = \frac{\sin \theta}{\cos \theta}$

The Quotient Identity is $\tan \theta = \frac{\sin \theta}{\cos \theta}$. We see that this is true because tangent is equal to $\frac{y}{x}$, in the unit circle. We know that $y$ is equal to the sine values of $\theta$ and $x$ is equal to the cosine values of $\theta$. Substituting $\sin \theta$ for $y$ and $\cos \theta$ for $x$ and we have a new identity.

Example 1: Use $\theta = 45^\circ$ to show that $\tan \theta = \frac{\sin \theta}{\cos \theta}$ holds true.

Solution: Plugging in $45^\circ$, we have: $\tan 45^\circ = \frac{\sin 45^\circ}{\cos 45^\circ}$. Then, substitute each function with its actual value and simplify both sides.

$\frac{\sin 45^\circ}{\cos 45^\circ} = \frac{\frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = \frac{\sqrt{2}}{2} \div \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{2} \cdot \frac{2}{\sqrt{2}} = 1$ and we know that tan $45^\circ = 1$, so this is true.

Example 2: Show that tan $90^\circ$ is undefined using the Quotient Identity.

Solution: $\tan 90^\circ = \frac{\sin 90^\circ}{\cos 90^\circ} = \frac{1}{0}$, because you cannot divide by zero, the tangent at $90^\circ$ is undefined.

Reciprocal Identities

Chapter 1 also introduced us to the idea that the three fundamental reciprocal trigonometric functions are cosecant (csc), secant (sec) and cotangent (cot) and are defined as:

$\csc \theta = \frac{1}{\sin \theta}\ \sec \theta = \frac{1}{\cos \theta}\ \cot \theta = \frac{1}{\tan \theta}$

If we apply the Quotient Identity to the reciprocal of tangent, an additional quotient is created:

$\cot \theta = \frac{1}{\tan \theta} = \frac{1}{\frac{\sin \theta}{\cos \theta}} = \frac{\cos \theta}{\sin \theta}$

Example 3: Prove $\tan \theta = \sin \theta \sec \theta$

Solution: First, you should change everything into sine and cosine. Feel free to work from either side, as long as the end result from both sides ends up being the same.

$\tan \theta & = \sin \theta \ \sec \theta \\& = \sin \theta \cdot \frac{1}{\cos \theta} \\& = \frac{\sin \theta}{\cos \theta}$

Here, we end up with the Quotient Identity, which we know is true. Therefore, this identity is also true and we are finished.

Example 4: Given $\sin \theta = - \frac{\sqrt{6}}{5}$ and $\theta$ is in the fourth quadrant, find $\sec \theta$.

Solution: Secant is the reciprocal of cosine, so we need to find the adjacent side. We are given the opposite side, $\sqrt{6}$ and the hypotenuse, $5$. Because $\theta$ is in the fourth quadrant, cosine will be positive. From the Pythagorean Theorem, the third side is:

$\left (\sqrt{6} \right )^2 + b^2 & = 5^2 \\6 + b^2 & = 25\\ b^2 & = 19 \\b & = \sqrt{19}$

From this we can now find $\cos \theta = \frac{\sqrt{19}}{5}$. Since secant is the reciprocal of cosine, $\sec \theta = \frac{5}{\sqrt{19}}$, or $\frac{5\sqrt{19}}{19}$.

Pythagorean Identity

Using the fundamental trig functions, sine and cosine and some basic algebra can reveal some interesting trigonometric relationships. Note when a trig function such as $\sin \theta$ is multiplied by itself, the mathematical convention is to write it as $\sin^2 \theta$. ($\sin \theta^2$ can be interpreted as the sine of the square of the angle, and is therefore avoided.)

$\sin^2 \theta = \frac{y^2}{r^2}$ and $\cos^2 \theta = \frac{x^2}{r^2}$ or $\sin^2 \theta + \cos^2 \theta = \frac{y^2}{r^2} + \frac{x^2}{r^2} = \frac{x^2+y^2}{r^r}$

Using the Pythagorean Theorem for the triangle above: $x^2+y^2=r^2$

Then, divide both sides by $r^2, \frac{x^2+y^2}{r^2} = \frac{r^2}{r^2} = 1$. So, because $\frac{x^2+y^2}{r^2} = 1, \sin^2 \theta + \cos^2 \theta$ also equals $1$. This is known as the Trigonometric Pythagorean Theorem or the Pythagorean Identity and is written $\sin^2 \theta + \cos^2 \theta = 1$. Alternative forms of the Theorem are: $1 + \cot^2 \theta = \csc^2 \theta$ and $\tan^2 \theta + 1 = \sec^2 \theta$. The second form is found by taking the original form and dividing each of the terms by $\sin^2 \theta$, while the third form is found by dividing all the terms of the first by $\cos^2 \theta$.

Example 5: Use $30^\circ$ to show that $\sin^2 \theta + \cos^2 \theta = 1$ holds true.

Solution: Plug in $30^\circ$ and find the values of $\sin 30^\circ$ and $\cos 30^\circ$.

$\sin^2 30^\circ & + \cos^2 30^ \circ \\\left (\frac{1}{2} \right )^2 & + \left (\frac{\sqrt{3}}{2} \right )^2 \\\frac{1}{4} & + \frac{3}{4} = 1$

Even and Odd Identities

Functions are even or odd depending on how the end behavior of the graphical representation looks. For example, $y = x^2$ is considered an even function because the ends of the parabola both point in the same direction and the parabola is symmetric about the $y-$axis. $y = x^3$ is considered an odd function for the opposite reason. The ends of a cubic function point in opposite directions and therefore the parabola is not symmetric about the $y-$axis. What about the trig functions? They do not have exponents to give us the even or odd clue (when the degree is even, a function is even, when the degree is odd, a function is odd).

$& \underline{\text{Even Function}} && \underline{\text{Odd Function}} \\& y = (-x)^2 = x^2 && y = (-x)^3 = -x^3$

Let’s consider sine. Start with $\sin(-x)$. Will it equal $\sin x$ or $-\sin x$? Plug in a couple of values to see.

$\sin(-30^\circ) & = \sin 330^\circ = - \frac{1}{2} = - \sin 30^\circ \\\sin (-135^\circ) & = \sin 225^\circ = - \frac{\sqrt{2}}{2} = - \sin 135^\circ$

From this we see that sine is odd. Therefore, $\sin(-x) = -\sin x$, for any value of $x$. For cosine, we will plug in a couple of values to determine if it’s even or odd.

$\cos(-30^\circ) & = \cos 330^\circ = \frac{\sqrt{3}}{2} = \cos 30^\circ \\\cos (-135^\circ) & = \cos 225^\circ = - \frac{\sqrt{2}}{2} = \cos 135^\circ$

This tells us that the cosine is even. Therefore, $\cos(-x) = \cos x$, for any value of $x$. The other four trigonometric functions are as follows:

$\tan(-x) & = - \tan x \\\csc (-x) & = - \csc x \\\sec(-x) & = \sec x \\\cot (-x) & = - \cot x$

Notice that cosecant is odd like sine and secant is even like cosine.

Example 6: If $\cos (-x) = \frac{3}{4}$ and $\tan(-x) = - \frac{\sqrt{7}}{3}$, find $\sin x$.

Solution: We know that sine is odd. Cosine is even, so $\cos x = \frac{3}{4}$. Tangent is odd, so $\tan x = \frac{\sqrt{7}}{3}$. Therefore, sine is positive and $\sin x = \frac{\sqrt{7}}{4}$.

Cofunction Identities

Recall that two angles are complementary if their sum is $90^\circ$. In every triangle, the sum of the interior angles is $180^\circ$ and the right angle has a measure of $90^\circ$. Therefore, the two remaining acute angles of the triangle have a sum equal to $90^\circ$, and are complementary. Let’s explore this concept to identify the relationship between a function of one angle and the function of its complement in any right triangle, or the cofunction identities. A cofunction is a pair of trigonometric functions that are equal when the variable in one function is the complement in the other.

In $\triangle{ABC}, \angle{C}$ is a right angle, $\angle{A}$ and $\angle{B}$ are complementary.

Chapter 1 introduced the cofunction identities (section 1.8) and because $\angle{A}$ and $\angle{B}$ are complementary, it was found that $\sin A = \cos B, \cos A = \sin B, \tan A = \cot B, \cot A = \tan B, \csc A = \sec B$ and $\sec A = \csc B$. For each of the above $\angle{A} = \frac{\pi}{2} - \angle {B}$. To generalize, $\sin \left (\frac{\pi}{2} - \theta \right ) = \cos \theta$ and $\cos \left (\frac{\pi}{2} - \theta\right ) = \sin \theta, \tan \left (\frac{\pi}{2} - \theta \right ) = \cot \theta$ and $\cot \left (\frac{\pi}{2} - \theta \right ) = \tan \theta, \csc \left (\frac{\pi}{2} - \theta \right ) = \sec \theta$ and $\sec \left (\frac{\pi}{2}- \theta \right ) = \csc \theta$.

The following graph represents two complete cycles of $y = \sin x$ and $y = \cos \theta$.

Notice that a phase shift of $\frac{\pi}{2}$ on $y = \cos x$, would make these graphs exactly the same. These cofunction identities hold true for all real numbers for which both sides of the equation are defined.

Example 7: Use the cofunction identities to evaluate each of the following expressions:

a. If $\tan \left (\frac{\pi}{2} - \theta \right )= - 4.26$ determine $\cot \theta$

b. If $\sin \theta = 0.91$ determine $\cos \left (\frac{\pi}{2} - \theta \right )$.

Solution:

a. $\tan \left (\frac{\pi}{2} - \theta \right ) = \cot \theta$ therefore $\cot \theta = -4.26$

b. $\cos \left (\frac{\pi}{2} - \theta \right ) = \sin \theta$ therefore $\cos \left (\frac{\pi}{2} - \theta \right ) = 0.91$

Example 8: Show $\sin \left (\frac{\pi}{2} - x \right ) = \cos (-x)$ is true.

Solution: Using the identities we have derived in this section, $\sin \left (\frac{\pi}{2} - x \right ) = \cos x$, and we know that cosine is an even function so $\cos(-x) = \cos x$. Therefore, each side is equal to $\cos x$ and thus equal to each other.

Points to Consider

• Why do you think secant is even like cosine?
• How could you show that tangent is odd?

Review Questions

1. Use the Quotient Identity to show that the tan $270^\circ$ is undefined.
2. If $\cos \left (\frac{\pi}{2} - x \right ) = \frac{4}{5}$, find $\sin(-x)$.
3. If $\tan(-x) = - \frac{5}{12}$ and $\sin x = - \frac{5}{13}$, find $\cos x$.
4. Simplify $\sec x \cos \left (\frac{\pi}{2} - x \right )$.
5. Verify $\sin^2 \theta + \cos^2 \theta = 1$ using:
1. the sides $5, 12$, and $13$ of a right triangle, in the first quadrant
2. the ratios from a $30-60-90$ triangle
6. Prove $1+ \tan^2 \theta = \sec^2 \theta$ using the Pythagorean Identity
7. If $\csc z = \frac{17}{8}$ and $\cos z= - \frac{15}{17}$, find $\cot z$.
8. Factor:
1. $\sin^2 \theta - \cos^2 \theta$
2. $\sin^2 \theta + 6 \sin \theta + 8$
9. Simplify $\frac{\sin^4 \theta - \cos^4 \theta}{\sin^2 \theta - \cos^2 \theta}$ using the trig identities
10. Rewrite $\frac{\cos x}{\sec x - 1}$ so that it is only in terms of cosine. Simplify completely.
11. Prove that tangent is an odd function.

1. $\tan 270^\circ = \frac{\sin 270^\circ}{\cos 270^\circ} = \frac{-1}{0}$, you cannot divide by zero, therefore $\tan 270^\circ$ is undefined.
2. If $\cos \left (\frac{\pi}{2} - x \right ) = \frac{4}{5}$, then, by the cofunction identities, $\sin x = \frac{4}{5}$. Because sine is odd, $\sin(-x) = - \frac{4}{5}$.
3. If $\tan(-x) = - \frac{5}{12}$, then $\tan x = \frac{5}{12}$. Because $\sin x = -\frac{5}{13}$, cosine is also negative, so $\cos x = - \frac{12}{13}$.
4. Use the reciprocal and cofunction identities to simplify $& \sec x \cos \left (\frac{\pi}{2} - x \right ) \\& \frac{1}{\cos x} \cdot \sin x \\& \frac{\sin x}{\cos x} \\& \tan x$
5. (a) Using the sides 5, 12, and 13 and in the first quadrant, it doesn’t really matter which is cosine or sine. So, $\sin^2 \theta + \cos^2 \theta = 1$ becomes $\left (\frac{5}{13} \right )^2 + \left (\frac{12}{13} \right )^2 = 1$. Simplifying, we get: $\frac{25}{169} + \frac{144}{169} = 1$, and finally $\frac{169}{169} = 1$. (b) $\sin^2 \theta + \cos^2 \theta = 1$ becomes $\left (\frac{1}{2} \right )^2 + \left (\frac{\sqrt{3}}{2} \right )^2 = 1$. Simplifying we get: $\frac{1}{4} + \frac{3}{4} = 1$ and $\frac{4}{4} = 1$.
6. To prove $\tan^2 \theta + 1 = \sec^2 \theta$, first use $\frac{\sin \theta}{\cos \theta} = \tan \theta$ and change $\sec^2 \theta = \frac{1}{\cos^2 \theta}$. $\tan^2 \theta + 1 & = \sec^2 \theta \\\frac{\sin^2 \theta}{\cos^2 \theta} + 1 & = \frac{1}{\cos^2 \theta} \\\frac{\sin^2 \theta}{\cos^2 \theta} + \frac{\cos^2 \theta}{\cos^2 \theta} & = \frac{1}{\cos^2 \theta} \\\sin^2 \theta + \cos^2 \theta & = 1$
7. If $\csc z = \frac{17}{8}$ and $\cos z = -\frac{15}{17}$, then $\sin z = \frac{8}{17}$ and $\tan z = - \frac{8}{15}$. Therefore $\cot z = - \frac{15}{8}$.
8. (a) Factor $\sin^2 \theta - \cos^2 \theta$ using the difference of squares. $\sin^2 \theta - \cos^2 \theta = (\sin \theta + \cos \theta) (\sin \theta - \cos \theta)$ (b) $\sin^2 \theta + 6\ \sin \theta + 8 = (\sin \theta + 4) (\sin \theta + 2)$
9. You will need to factor and use the $\sin^2 \theta + \cos^2 \theta = 1$ identity. $& \frac{\sin^4 \theta - \cos^4 \theta}{\sin^2 \theta - \cos^2 \theta} \\& \qquad \quad \ \ = \frac{(\sin^2 \theta - \cos^2 \theta)(\sin^2 \theta + \cos^2 \theta)}{\sin^2 \theta - \cos^2 \theta} \\& \qquad \quad \ \ = \sin^2 \theta + \cos^2 \theta \\& \qquad \quad \ \ = 1$
10. To rewrite $\frac{\cos x}{\sec x -1}$ so it is only in terms of cosine, start with changing secant to cosine. $\frac{\cos x}{\sec x-1} & = \frac{\cos x}{\frac{1}{\cos x}-1} && \text{Now, simplify the denominator}. \\\frac{\cos x}{\frac{1}{\cos x}-1} & = \frac{\cos x}{\frac{1-\cos x}{\cos x}}$ Multiply by the reciprocal $\frac{\cos x}{\frac{1- \cos x}{\cos x}} = \cos x \div \frac{1- \cos x}{\cos x} = \cos x \cdot \frac{\cos x}{1- \cos x} = \frac{\cos^2 x}{1-\cos x}$
11. The easiest way to prove that tangent is odd to break it down, using the Quotient Identity. $\tan(-x)& =\frac{\sin(-x)}{\cos (-x)}&& \text{from this statement, we need to show that}\ \tan(-x) = - \tan x \\& = \frac{- \sin x}{\cos x} && \text{because}\ \sin(-x) = - \sin x\ \text{and}\ \cos(-x) = \cos x \\& = - \tan x$

Feb 23, 2012

Dec 16, 2014