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Trigonometric Identities

Trigonometric Identities

Assessment

Presentation

Mathematics

12th Grade

Hard

Created by

Teacher Shivam

Used 1+ times

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5 Slides • 0 Questions

1

​Identities

Trigonometric Identities​

2

​Circumference of a circle = 2πr

π​ = 3.14159 (22∕7 or 355/113)

​A radian = 57° 17′ 44.8″

Angle = (arc/radius) ×​ Radian

sin²​θ + cos²θ = 1

sec²​θ = 1 + tan²θ

​cosec²θ = 1 + cot²θ

sin 0⁰​ = 0 and cos 0⁰ = 1

sin 30⁰​ = 1/2 and cos 30⁰ = √3 / 2

sin 45⁰​ = cos 45⁰ = 1

​sin 60⁰​ = √3 / 2 and cos 60⁰ = 1/2

sin 15⁰​ = (√3 - 1) / 2√2 and cos 15⁰ = (√3 + 1) / 2√2

3

​sin (-θ) = - sin θ; cos (-θ​) = cos θ

sin (90⁰​ - θ) = cos θ; cos (90⁰ - θ) = sin θ

sin (90⁰​ + θ) = cos θ; cos (90 + θ) = -sin θ

sin (180⁰ - θ​) = sin θ; cos (180⁰ - θ) = -cos θ

sin (180⁰​ + θ) = -sin θ; cos (180⁰ + θ) = -cos θ

If sin θ​ = sin α, then θ = nπ ± α

If cos θ​ = cos α, then θ = 2nπ ± α

​If tan θ = tan α, then θ = nπ + α

4

​sin (A + B) = sin A cos B + cos A sin B

cos (A + B) = ​cos A cos B - sin A sin B

​sin (A - B) = sin A cos B - cos A sin B

cos (A - B) = cos A cos B + sin A sin B

sin C + sin D = 2 . sin (C+D)/2 . cos (C-D)/2

sin C - sin D = 2 . cos (C+D)/2 . sin (C-D)/2

cos C + cos D = 2 . cos (C+D)/2 . cos (C-D)/2​

​2 sin A cos B = sin (A + B) + sin (A - B)

2 cos A sin B = sin (A + B) - sin (A - B)​

​2 cos A cos B = cos (A + B) + cos (A - B)

2 sin A sin B = cos (A - B) - cos (A + B)

5

​tan (A + B) = (tan A + tan B) / (1 - tan A tan B)

tan (A - B) = (tan A - tan B) / (1 + tan A tan B) ​

sin 2A = 2 sin A cos A

cos 2A = cos​² A - sin² A = 1 - 2 sin² A = 2 cos² A - 1

sin 2A = (2 tan A) / (1 + tan²​ A); cos 2A = (1 - tan² A) / (1 + tan² A)

tan 2A = 2tan A / (1 - tan²​ A)

sin​ 3A = 3 sin A - 4 sin³ A

​cos 3A = 4 cos³ A - 3 cos A

tan 3A = (3 tan A - tan³​ A) / (1 - 3tan² A)

​sin (A / 2) = ± √ ((1 - cos A) / 2); cos (A / 2) = ± √ ((1 + cos A) / 2)

​2 sin (A / 2) = ± √ (1 + sin A) ± √ ( 1 - sin A)

2 cos (A / 2) = ± √ (1 + sin A) ∓ √ (1 - sin A)

​Identities

Trigonometric Identities​

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