Pythagorean Identities

Pythagorean Identities in trigonometry are derived from the Pythagorean Theorem. They are also called Pythagorean Trigonometric Identities.

The three basic Pythagorean identities are:

However, from these identities, we can derive the other six identities as well.

All 9 Pythagorean identities are:

Pythagorean Theorem

The Pythagorean theorem states that in a right-angled triangle, the square of the hypotenuse is equal to the sum of the squares of the perpendicular and base. The formula for the Pythagorean theorem is given by:

(Hypotenuse)² = (Perpendicular)² + (Base)²

The above figure is of a right-angled triangle ABC with perpendicular AB, base BC, and hypotenuse AC. So, by the Pythagorean theorem: AC² = AB² + BC²

Pythagorean Trigonometric Identities: Derivation

Pythagorean identities can be easily proved by applying the Pythagorean theorem in a right-angled triangle. Let's discuss the proof of each identity in detail as follows:

Consider a right-angled triangle ABC with perpendicular AB, base BC, and hypotenuse AC.

Derivation of sin²θ + cos²θ = 1

By the Pythagorean theorem: AC² = AB² + BC²

Dividing both sides by AC²

[AC² / AC²] = [AB² / AC²] + [BC² / AC²]

⇒ 1 = [AB / AC]² + [BC / AC]² . . . (a)

From the above triangle

[AB / AC] = sinθ, and [BC / AC] = cosθ

Putting these values in equation (a), we get the following:

⇒ sin²θ + cos²θ = 1

Hence Proved.

Derivation of 1 + tan²θ = sec²θ

By the Pythagorean theorem, AC² = AB² + BC²

Dividing both sides by BC²

[AC² / BC²] = [AB² / BC²] + [BC² / BC²]

⇒ [AC / BC]² = [AB / BC]² + 1 . . . (b)

From the above triangle, we have

[AC / BC] = sec θ, and

[AB / BC] = tan θ

Putting these values in equation (b), we get the following:

⇒ sec²θ = 1 + tan²θ

Hence Proved

Derivation of 1 + cot²θ = cosec²θ

By the Pythagorean theorem, we have AC² = AB² + BC²

Dividing both by AB², we get

[AC² / AB²] = [AB² / AB²] + [BC² / AB²]

⇒ [AC / AB]² = 1 + [BC / AB] . . . (c)

From the above triangle, we know

[AC / AB] = cosec θ, and [BC / AB] = cot θ

Putting these values in equation (c), we get:

⇒ cosec²θ = 1 + cot²θ

Hence Proved

Some of the other identities include:

• Sum and Difference Identities
• Half-Angle Identity
• Double Angle Identity

Related Articles:

• Trigonometry
• Trigonometric Formulas
• Pythagorean Triples
• Applications of Trigonometry

Solved Examples

Example 1: If the angle y is in the second quadrant and cos y = -5/13. Find the value of sin y.

Solution:

By Pythagorean identity

sin²y + cos²y = 1
⇒ sin²y + (-5/13)²= 1
⇒ sin²y = 1 - (25/169)
⇒ sin²y = 144 / 169
⇒ sin y = ± 12/13

Since, sin is positive in second quadrant so we will take positive value i.e., sin y = 12/13.

Example 2: Evaluate [1/ (1 - cos x)] + [1/ (1 + cos x)]

Solution:

Given, [1/ (1 - cos x)] + [1/ (1 + cos x)]

= [(1 + cosx) + (1 - cos x)] / [(1 + cosx) (1 - cos x)]
= 2 / (1²- cos²x)
= 2 / (1 - cos²x)

By Pythagorean identity, sin²y + cos²y = 1 or sin²y = 1 - cos²y

⇒ [1/ (1 - cos x)] + [1/ (1 + cos x)] = 2 / sin²x
⇒ [1/ (1 - cos x)] + [1/ (1 + cos x)] = 2 cosec²x

Example 3: Simplify (tan θ + sec θ)(tan θ - sec θ)

Solution:

Given, (tan θ + sec θ)(tan θ - sec θ)

= (tan² θ - sec² θ) [ As (a - b)(a + b) = a² - b²]

By Pythagorean identity, we have

sec² θ - tan² θ = 1
(tan θ + sec θ)(tan θ - sec θ) = -(sec² θ - tan² θ) = -1

Hence, (tan θ + sec θ)(tan θ - sec θ) = -1

Example 4: Evaluate (cosec z + cot z) [(1 - cos z)/sin z]

Solution:

Given, (cosec z + cot z) [(1 - cos z)/sin z]

Simplifying,

(cosec z + cot z) [(1/sin z) - (cos z/sin z)]
= (cosec z + cot z) (cosec z - cot z) [As 1/ sin z = cosec z and cos z/sinz = cot z]
= (cosec²z - cot²z) [ As (a-b)(a+b) = a² - b²]

By Pythagorean Identity we have, cosec²θ - cot²θ = 1

Hence, (cosec z + cot z) [(1 - cos z)/sin z] = 1

Practice Questions

Question 1: If cot A = 4/3 then, find the value of cosec A.

Question 2: Evaluate [1/ (1 - cosec x)] [1/ (1 + cosec x)].

Question 3: Simplify (cosec θ + cot θ)(cosec θ - cot θ).

Question 4: Prove that sinx - cos⁴x = sin²x - cos²x.