
Trigonometric Identities
Presentation
•
Mathematics
•
12th Grade
•
Hard
Teacher Shivam
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5 Slides • 0 Questions
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Trigonometric Identities
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Circumference of a circle = 2πr
π = 3.14159 (22∕7 or 355/113)
A radian = 57° 17′ 44.8″
Angle = (arc/radius) × Radian
sin²θ + cos²θ = 1
sec²θ = 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)
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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 = cos² 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)
Trigonometric Identities
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