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Functions and Their Modeling

Trigonometric Identity

Understanding Trigonometric Identities

Have you ever noticed that some mathematical equations are always true for any value of their variables? For example, (a+b)2=a2+2ab+b2(a + b)^2 = a^2 + 2ab + b^2 is always true for any values of a and b. Equations like this are called identities.

In trigonometry, we also have equations that are always true for any angle value. These are called trigonometric identities. These identities are very useful for simplifying trigonometric expressions and solving equations.

Basic Trigonometric Identities

Pythagorean Identity

Let's start with the most fundamental identity. Consider a unit circle with point P(x,y)P(x, y) that forms angle θ\theta with the positive x-axis.

Unit Circle and Pythagorean Identity
Notice how the coordinates of point P change as the angle changes. These coordinates are the values of cos θ and sin θ.
0.79 Radian
0°360°

On the unit circle:

  • Radius = 1
  • x-coordinate = cos⁡θ\cos \theta
  • y-coordinate = sin⁡θ\sin \theta

Using the Pythagorean theorem for point P:

x2+y2=12x^2 + y^2 = 1^2

Substituting the values of x and y:

(cos⁡θ)2+(sin⁡θ)2=1(\cos \theta)^2 + (\sin \theta)^2 = 1

Or can be written as:

sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1

This is the Pythagorean identity, the most fundamental identity in trigonometry.

Other Forms of Pythagorean Identity:

From the basic identity above, we can derive two other forms:

Second form: Divide both sides by cos⁡2θ\cos^2 \theta (for cos⁡θ≠0\cos \theta \neq 0)

sin⁡2θcos⁡2θ+cos⁡2θcos⁡2θ=1cos⁡2θ\frac{\sin^2 \theta}{\cos^2 \theta} + \frac{\cos^2 \theta}{\cos^2 \theta} = \frac{1}{\cos^2 \theta}
tan⁡2θ+1=sec⁡2θ\tan^2 \theta + 1 = \sec^2 \theta

Third form: Divide both sides by sin⁡2θ\sin^2 \theta (for sin⁡θ≠0\sin \theta \neq 0)

sin⁡2θsin⁡2θ+cos⁡2θsin⁡2θ=1sin⁡2θ\frac{\sin^2 \theta}{\sin^2 \theta} + \frac{\cos^2 \theta}{\sin^2 \theta} = \frac{1}{\sin^2 \theta}
1+cot⁡2θ=csc⁡2θ1 + \cot^2 \theta = \csc^2 \theta

Reciprocal Identities

Each trigonometric function has its reciprocal. This relationship forms reciprocal identities:

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

Or in the opposite form:

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

Quotient Identities

Quotient identities relate tangent and cotangent to sine and cosine:

tan⁡θ=sin⁡θcos⁡θ\tan \theta = \frac{\sin \theta}{\cos \theta}
cot⁡θ=cos⁡θsin⁡θ\cot \theta = \frac{\cos \theta}{\sin \theta}
Trigonometric Function Visualization
Observe how the ratios of triangle sides give sin, cos, and tan values. Also notice how these values change as the angle changes.
Sin (30°) = 0.50Cos (30°) = 0.87Tan (30°) = 0.58
0.52 Radian
0°360°

Both identities can be proven directly from the definition of trigonometric functions on the unit circle.

Even and Odd Function Identities

When angles are negative, trigonometric functions have special properties:

Even function (symmetry about y-axis):

cos⁡(−θ)=cos⁡θ\cos(-\theta) = \cos \theta

Odd functions (symmetry about origin):

sin⁡(−θ)=−sin⁡θ\sin(-\theta) = -\sin \theta
tan⁡(−θ)=−tan⁡θ\tan(-\theta) = -\tan \theta
Exploring Even and Odd Properties
Try moving the slider to negative and positive values. Notice how cos, sin, and tan values change for opposite angles.
1.05 Radian
0°360°

Using Identities in Proofs

Let's see how trigonometric identities are used to prove other equations.

Simplifying Expressions

Simplify sin⁡2θcos⁡θ+cos⁡θ\frac{\sin^2 \theta}{\cos \theta} + \cos \theta

Solution:

sin⁡2θcos⁡θ+cos⁡θ=sin⁡2θcos⁡θ+cos⁡2θcos⁡θ\frac{\sin^2 \theta}{\cos \theta} + \cos \theta = \frac{\sin^2 \theta}{\cos \theta} + \frac{\cos^2 \theta}{\cos \theta}
=sin⁡2θ+cos⁡2θcos⁡θ= \frac{\sin^2 \theta + \cos^2 \theta}{\cos \theta}
=1cos⁡θ(using Pythagorean identity)= \frac{1}{\cos \theta} \quad \text{(using Pythagorean identity)}
=sec⁡θ= \sec \theta

Proving Identities

Prove that 1+tan⁡2θsec⁡θ=sec⁡θ\frac{1 + \tan^2 \theta}{\sec \theta} = \sec \theta

Solution:

We start from the left side:

1+tan⁡2θsec⁡θ=sec⁡2θsec⁡θ(using 1+tan⁡2θ=sec⁡2θ)\frac{1 + \tan^2 \theta}{\sec \theta} = \frac{\sec^2 \theta}{\sec \theta} \quad \text{(using } 1 + \tan^2 \theta = \sec^2 \theta \text{)}
=sec⁡θ= \sec \theta

It is proven that the left side equals the right side.

Determining Trigonometric Function Values

Trigonometric identities are very useful for determining the values of all trigonometric functions when one of them is known.

Identity Applications

If sin⁡θ=35\sin \theta = \frac{3}{5} and 90°<θ<180°90° < \theta < 180° (quadrant II), determine the values of other trigonometric functions.

Solution:

Use the Pythagorean identity to find cos⁡θ\cos \theta:

sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1
(35)2+cos⁡2θ=1\left(\frac{3}{5}\right)^2 + \cos^2 \theta = 1
925+cos⁡2θ=1\frac{9}{25} + \cos^2 \theta = 1
cos⁡2θ=1−925=1625\cos^2 \theta = 1 - \frac{9}{25} = \frac{16}{25}
cos⁡θ=±45\cos \theta = \pm \frac{4}{5}

Since θ\theta is in quadrant II, then cos⁡θ<0\cos \theta < 0. Therefore, cos⁡θ=−45\cos \theta = -\frac{4}{5}

Next, calculate the other trigonometric functions:

tan⁡θ=sin⁡θcos⁡θ=3/5−4/5=−34\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{3/5}{-4/5} = -\frac{3}{4}
csc⁡θ=1sin⁡θ=53\csc \theta = \frac{1}{\sin \theta} = \frac{5}{3}
sec⁡θ=1cos⁡θ=−54\sec \theta = \frac{1}{\cos \theta} = -\frac{5}{4}
cot⁡θ=1tan⁡θ=−43\cot \theta = \frac{1}{\tan \theta} = -\frac{4}{3}

Exercises

  1. Simplify the expression tan⁡θ⋅cos⁡θsin⁡θ\frac{\tan \theta \cdot \cos \theta}{\sin \theta}

  2. Prove the identity sin⁡θ1+cos⁡θ=1−cos⁡θsin⁡θ\frac{\sin \theta}{1 + \cos \theta} = \frac{1 - \cos \theta}{\sin \theta}

  3. If cos⁡θ=513\cos \theta = \frac{5}{13} and 270°<θ<360°270° < \theta < 360°, determine the values of all trigonometric functions.

  4. Simplify sin⁡4θ−cos⁡4θ\sin^4 \theta - \cos^4 \theta

  5. If tan⁡θ=43\tan \theta = \frac{4}{3} and sin⁡θ<0\sin \theta < 0, determine the values of sin⁡θ\sin \theta and cos⁡θ\cos \theta.

Answer Key

  1. Let's simplify step by step:

    tan⁡θ⋅cos⁡θsin⁡θ=sin⁡θcos⁡θ⋅cos⁡θsin⁡θ\frac{\tan \theta \cdot \cos \theta}{\sin \theta} = \frac{\frac{\sin \theta}{\cos \theta} \cdot \cos \theta}{\sin \theta}
    =sin⁡θsin⁡θ=1= \frac{\sin \theta}{\sin \theta} = 1

  2. To prove the identity, we will transform the left side:

    sin⁡θ1+cos⁡θ=sin⁡θ1+cos⁡θ⋅1−cos⁡θ1−cos⁡θ\frac{\sin \theta}{1 + \cos \theta} = \frac{\sin \theta}{1 + \cos \theta} \cdot \frac{1 - \cos \theta}{1 - \cos \theta}
    =sin⁡θ(1−cos⁡θ)(1+cos⁡θ)(1−cos⁡θ)= \frac{\sin \theta (1 - \cos \theta)}{(1 + \cos \theta)(1 - \cos \theta)}
    =sin⁡θ(1−cos⁡θ)1−cos⁡2θ= \frac{\sin \theta (1 - \cos \theta)}{1 - \cos^2 \theta}
    =sin⁡θ(1−cos⁡θ)sin⁡2θ= \frac{\sin \theta (1 - \cos \theta)}{\sin^2 \theta}
    =1−cos⁡θsin⁡θ= \frac{1 - \cos \theta}{\sin \theta}

  3. Given cos⁡θ=513\cos \theta = \frac{5}{13} in quadrant IV.

    Finding sin⁡θ\sin \theta:

    sin⁡2θ=1−cos⁡2θ=1−25169=144169\sin^2 \theta = 1 - \cos^2 \theta = 1 - \frac{25}{169} = \frac{144}{169}
    sin⁡θ=−1213 (negative in quadrant IV)\sin \theta = -\frac{12}{13} \text{ (negative in quadrant IV)}

    Other trigonometric functions:

    tan⁡θ=−12/135/13=−125\tan \theta = \frac{-12/13}{5/13} = -\frac{12}{5}
    csc⁡θ=−1312\csc \theta = -\frac{13}{12}
    sec⁡θ=135\sec \theta = \frac{13}{5}
    cot⁡θ=−512\cot \theta = -\frac{5}{12}

  4. Use difference of squares factoring:

    sin⁡4θ−cos⁡4θ=(sin⁡2θ)2−(cos⁡2θ)2\sin^4 \theta - \cos^4 \theta = (\sin^2 \theta)^2 - (\cos^2 \theta)^2
    =(sin⁡2θ+cos⁡2θ)(sin⁡2θ−cos⁡2θ)= (\sin^2 \theta + \cos^2 \theta)(\sin^2 \theta - \cos^2 \theta)
    =1⋅(sin⁡2θ−cos⁡2θ)= 1 \cdot (\sin^2 \theta - \cos^2 \theta)
    =sin⁡2θ−cos⁡2θ= \sin^2 \theta - \cos^2 \theta

  5. Given tan⁡θ=43\tan \theta = \frac{4}{3} and sin⁡θ<0\sin \theta < 0.

    Since tan⁡θ>0\tan \theta > 0 and sin⁡θ<0\sin \theta < 0, then cos⁡θ<0\cos \theta < 0 (quadrant III).

    Use the identity 1+tan⁡2θ=sec⁡2θ1 + \tan^2 \theta = \sec^2 \theta:

    1+169=sec⁡2θ1 + \frac{16}{9} = \sec^2 \theta
    sec⁡2θ=259\sec^2 \theta = \frac{25}{9}
    sec⁡θ=−53 (negative in quadrant III)\sec \theta = -\frac{5}{3} \text{ (negative in quadrant III)}
    cos⁡θ=−35\cos \theta = -\frac{3}{5}

    For sin⁡θ\sin \theta:

    tan⁡θ=sin⁡θcos⁡θ\tan \theta = \frac{\sin \theta}{\cos \theta}
    43=sin⁡θ−3/5\frac{4}{3} = \frac{\sin \theta}{-3/5}
    sin⁡θ=43⋅(−35)=−45\sin \theta = \frac{4}{3} \cdot \left(-\frac{3}{5}\right) = -\frac{4}{5}
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