Table of Contents

Last modified on August 3rd, 2023

An identity is an equation that is true for all possible values that are substituted in the equation. A Pythagorean identity, also known as the Pythagorean trigonometric identity, is an identity expressing the Pythagorean Theorem in terms of trigonometric functions.

The 3 Pythagorean trigonometric identities are shown below:

It is the most basic or fundamental Pythagorean identity and is given by the expression:

**sin ^{2 }θ + cos^{2 }θ = 1**

Pythagorean identities are useful in simplifying trigonometric expressions having trigonometric functions such as sin, cos, and tan.

Let us learn how to derive the fundamental Pythagorean identity.

Consider a right triangle ABC with side lengths a, b, and c that follows the Pythagorean Theorem

a^{2} + b^{2} = c^{2}, here a & b are the two legs, c = hypotenuse

Dividing both sides of the expression by c^{2}, we get,

=> a^{2}/c^{2} + b^{2}/c^{2} = /c^{2}/c^{2}

=> (a/c)^{2} + (b/c)^{2} = 1

=> cos^{2 }θ + sin^{2 }θ = 1 (∵ a/c = base/hypotenuse = cos θ, b/c = perpendicular/hypotenuse = sin θ)

=> **sin ^{2 }θ + cos^{2 }θ = 1**

The above identity is thus another form of the Pythagorean Theorem and is true for all values of θ

**Alternative Method – Using Unit Circle**

The above Pythagorean identity can also be obtained using a unit circle, a circle of radius 1 unit (c = 1) we learned that a point on the unit circle is represented by the coordinates (a, b) = (cos θ, sinθ). Thus, the height or perpendicular of the right triangle is sinθ and the base is cos θ as shown below:

If we place the right triangle ABC in a unit circle and apply the Pythagorean Theorem, we get

**a ^{2} + b^{2} = c^{2}**

=> **sin ^{2 }θ + cos^{2 }θ = 1** (∵ a = cos θ, b = sin θ)

Using the above Pythagorean identity, we can obtain 2 more Pythagorean identities. Let us see how we can obtain them.

As we know, fundamental Pythagorean identity is given by:

cos^{2 }θ + sin^{2 }θ = 1

Dividing both sides of the equation by cos^{2 }θ, we get

=> cos^{2 }θ/ cos^{2 }θ + sin^{2 }θ/ cos^{2 }θ = 1/ cos^{2 }θ

=> **1 + tan ^{2 }θ = sec^{2 }θ** (∵ sin

The above equation is the secondPythagorean identity. Here, it should be noted that there are values of θ for which tangent and secant is undefined:

When cos θ = 0, tan θ = sin θ/cos and sec θ = 1/cos θ are undefined

Similarly, the third Pythagorean identity is obtained as shown below:

Here, we will again use the fundamental Pythagorean identity

cos^{2 }θ + sin^{2 }θ = 1

Dividing both sides of the equation by sin^{2 }θ, we get

=> cos^{2 }θ/ sin^{2 }θ + sin^{2 }θ/ sin^{2 }θ = 1/ sin^{2 }θ

=> cot^{2 }θ + 1 = cosec^{2 }θ

=> **1 + cot ^{2 }θ = cosec^{2 }θ** (∵ cos

When sin θ = 0, cot θ = cos θ/sin and cosec θ = 1/sin θ are undefined

A list of allthePythagorean identities and their variations are given in the table below:

Pythagorean Identities | Variations |
---|---|

sin^{2 }θ + cos^{2 }θ = 1 | sin^{2 }θ = 1 – cos^{2 }θ, cos^{2 }θ = 1 – sin^{2 }θ |

1 + tan^{2 }θ = sec^{2 }θ | tan^{2 }θ = sec^{2 }θ – 1 |

1 + cot^{2 }θ = cosec^{2 }θ | cot^{2 }θ = cosec^{2 }θ – 1 |

Let us learn how to use Pythagorean identities involving solved examples.

**If sin 30° = ½. Find the value of cos 30°.**

Solution:

As we know,

sin^{2 }θ + cos^{2 }θ = 1

=> sin^{2 }30°+ cos^{2 }30° = 1

=> (1/2)^{2} + cos^{2 }(30°) = 1 (∵ sin (30°) = ½)

=> cos^{2 }(30°) = 1 – ¼

=> cos^{2 }(30°) = ¾

=> cos(30°) = √¾ = ±√3/2

Since 30° is an acute angle, the value of cos (30°)** = **√3/2

**If θ is in the second quadrant, and sin θ = 3/5, what is the value of cos θ?**

Solution:

Substituting the value of sin θ, and solving for cos θ, we get

sin^{2 }θ + cos^{2 }θ = 1

=> (3/5)^{2} + cos^{2 }θ = 1

=> cos^{2 }θ = 1 – (3/5)^{2}

=> cos^{2 }θ = 25/25 – 9/25

=> cos^{2 }θ = 16/25

=> cosθ = √16/25

=> cosθ = ±4/5

Since the cosine is in the second quadrant, the value of cosθ is -4/5

**Simplify the expression: (1 + cos θ)(1 – cos θ)**

Solution:

Given,

(1 + cos θ)(1 – cos θ)

=> 1 – cos^{2 }θ

=> sin^{2 }θ

Last modified on August 3rd, 2023

example #2 is wrong. Since it is in the 2nd quadrant, the cosine value should be -4/5, not positive. Quadrants 2 and 3 always give a negative cosine value as cosine represents x on the coordinate place.

Correction done

Really helpful and understandable