Pythagorean Identities in trigonometry are derived from the Pythagorean Theorem. They are also called Pythagorean Trigonometric Identities. One of the basic Pythagorean identities is sin²θ + cos²θ = 1. Let's learn more about, Pythagorean identities in trigonometry including their proof and solved examples, as well.
Pythagorean identities are important identities in trigonometry that are based on the Pythagoras theorem. It relates the square of one trigonometric ratio with the other. It can be used to solve complex trigonometric problems easily and also used to prove various other trigonometric identities.
List all Pythagorean Identities.
In trigonometry, we have three basic Pythagorean identities which are:
- sin2θ + cos2θ = 1
- 1 + tan2θ = sec2θ
- 1 + cot2θ = cosec2θ
However, from these identities, we can derive the other six identities as well.
All 9 Pythagorean identities are:
| Pythagorean Identity | Alternative Identites |
|---|
| sin2θ + cos2θ = 1 | 1 - sin2θ = cos2 θ
OR
1 - cos2θ = sin2θ |
| sec2θ - tan2θ = 1 | 1 + tan2θ = sec2θ
OR
sec2θ - 1 = tan2θ |
| cosec2θ - cot2θ = 1 | 1 + cot2θ = cosec2θ
OR
cosec2θ - 1 = cot2θ |
Pythagoras Theorem
Pythagoras theorem states that in a right-angled triangle, the square of hypotenuse is equal to the sum of squares of perpendicular and base. The formula for Pythagoras 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 Pythagoras 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 Pythagoras 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:
⇒ sin2θ + cos2θ = 1
Hence Proved.
Derivation of 1 + tan2θ = sec2θ
By the Pythagoras 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:
⇒ sec2θ = 1 + tan2θ
Hence Proved
Derivation of 1 + cot2θ = cosec2θ
By the Pythagoras theorem, we have AC2 = AB2 + BC2
Dividing both by AB2, we get
[AC2 / AB2] = [AB2 / AB2] + [BC2 / AB2]
⇒ [AC / AB]2 = 1 + [BC / AB]2 . . . (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
Applications of Pythagorean Identities
Some applications of Pythagorean identities in trigonometry are:
- Pythagorean identities can be used to derive other trigonometric identities.
- It can also be used in the heights and distances.
- It can be used to simplify trigonometric problems.
Check: Applications of Trigonometry
Other Trigonometric Identities
Some of the other identities include:
Read More,
Solved Examples of Pythagorean Identities
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 (tanθ + secθ)(tanθ - secθ)
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 on Pythagorean Identities
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 (cosecθ + cotθ)(cosecθ - cotθ).
Question 4: Prove that: sin4x - cos4x = sin2x - cos2x.
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