Trigonometric Formula Sheet, Exams of Trigonometry

Download Trigonometric Formula Sheet and more Exams Trigonometry in PDF only on Docsity! 1 2 2 2 (x, y) 1 y θ x Trigonometric Formula Sheet Definition of the Trig Functions Right Triangle Definition Assume that: Unit Circle Definition Assume θ can be any angle. 0 < θ < π or 0◦ < θ < 90◦ y hypotenuse θ adjacent opposite x sin θ = opp hyp csc θ = hyp opp sin θ = y 1 csc θ = 1 y cos θ = adj hyp sec θ = hyp adj cos θ = x 1 sec θ = 1 x tan θ = opp adj cot θ = adj opp tan θ = y x cot θ = x y sin θ, ∀ θ ∈ (−∞, ∞) cos θ, ∀ θ ∈ (−∞, ∞) Domains of the Trig Functions csc θ, ∀ θ /= nπ, where n ∈ Z sec θ, ∀ θ n + 1 π, where n ∈ Z tan θ, ∀ θ n + 1 π, where n ∈ Z cot θ, ∀ θ nπ, where n ∈ Z Ranges of the Trig Functions −1 ≤ sin θ ≤ 1 −1 ≤ cos θ ≤ 1 −∞ ≤ tan θ ≤ ∞ csc θ ≥ 1 and csc θ ≤ −1 sec θ ≥ 1 and sec θ ≤ −1 −∞ ≤ cot θ ≤ ∞ Periods of the Trig Functions The period of a function is the number, T, such that f (θ +T ) = f (θ ) . So, if ω is a fixed number and θ is any angle we have the following periods. 2πsin(ωθ) ⇒ T = ω 2πcos(ωθ) ⇒ T = ω π 2πcsc(ωθ) ⇒ T = ω 2πsec(ωθ) ⇒ T = ω π tan(ωθ) ⇒ T = ω cot(ωθ) ⇒ T = ω r 2 ± 2 2 2 2 2 2 2 2 2 2 2 Identities and Formulas Tangent and Cotangent Identities Half Angle Formulas sin θ tan θ = cos θ cos θ cot θ = sin θ sin θ = ± 1 − cos(2θ) 2 Reciprocal Identities cos θ = ± r 1 + cos(2θ) s 1 − cos(2 θ ) Pythagorean Identities sin2 θ + cos2 θ = 1 tan2 θ + 1 = sec2 θ 1 + cot2 θ = csc2 θ Even and Odd Formulas Sum and Difference Formulas sin(α ± β) = sin α cos β ± cos α sin β cos(α ± β) = cos α cos β ∓ sin α sin β tan(α β) = tan α ± tan β 1 ∓ tan α tan β Product to Sum Formulas 1 sin(−θ) = − sin θ cos(−θ) = cos θ tan(−θ) = − tan θ Periodic Formulas If n is an integer sin(θ + 2πn) = sin θ cos(θ + 2πn) = cos θ tan(θ + πn) = tan θ csc(−θ) = − csc θ sec(−θ) = sec θ cot(−θ) = − cot θ csc(θ + 2πn) = csc θ sec(θ + 2πn) = sec θ cot(θ + πn) = cot θ sin α sin β = 2 [cos(α − β) − cos(α + β)] 1 cos α cos β = 2 [cos(α − β) + cos(α + β)] 1 sin α cos β = 2 [sin(α + β) + sin(α − β)] 1 cos α sin β = 2 [sin(α + β) − sin(α − β)] Sum to Product Formulas sin α + sin β = 2 sin α + β cos α − β sin(2θ) = 2 sin θ cos θ cos(2θ) = cos2 θ − sin2 θ sin α − sin β = 2 cos α + β sin α − β cos α + cos β = 2 cos α + β cos α − β = 2 cos2 θ − 1 = 1 − 2 sin2 θ 2 tan θ cos α − cos β = −2 sin 2 α + β 2 sin 2 α − β 2 tan(2θ) = 1 − tan2 θ Degrees to Radians Formulas If x is an angle in degrees and t is an angle in radians then: Cofunction Formulas sin π − θ = cos θ csc π − θ = sec θ cos π − θ = sin θ sec π − θ = csc θ π t = πx ⇒ t = 180◦t and x = tan π − θ = cot θ tan θ = ± 1 + cos(2θ) Double Angle Formulas 180◦ x 180◦ π sin θ = 1 csc θ csc θ = 1 sin θ cos θ = 1 sec θ sec θ = 1 cos θ tan θ = 1 cot θ cot θ = 1 tan θ 2 2 2 2 = = = 2 5 β a c γ α Inverse Trig Functions Definition θ = sin−1(x) is equivalent to x = sin θ Inverse Properties These properties hold for x in the domain and θ in the range θ = cos−1(x) is equivalent to x = cos θ θ = tan−1(x) is equivalent to x = tan θ Domain and Range sin(sin−1(x)) = x cos(cos−1(x)) = x tan(tan−1(x)) = x sin−1(sin(θ)) = θ cos−1(cos(θ)) = θ tan−1(tan(θ)) = θ Function θ = sin−1(x) θ = cos−1(x) θ = tan−1(x) Domain −1 ≤ x ≤ 1 −1 ≤ x ≤ 1 −∞ ≤ x ≤ ∞ Range π π — 2 ≤ θ ≤ 2 0 ≤ θ ≤ π π π — < θ < Other Notations sin−1(x) = arcsin(x) cos−1(x) = arccos(x) 2 2 tan−1(x) = arctan(x) Law of Sines, Cosines, and Tangents b Law of Sines Law of Tangents sin α sin β sin γ a − b tan 1 (α − β) a b c Law of Cosines a + b b − c tan 1 (α + β) tan 1 (β − γ) a2 = b2 + c2 − 2bc cos α = b + c 2 tan 1 (β + γ) b2 = a2 + c2 − 2ac cos β a − c = tan 1 (α − γ) c2 = a2 + b2 − 2ab cos γ a + c 6 tan 1 (α + γ) 1 4 7 √ −a = i √ a, a ≥ 0 Complex Numbers i = √ −1 i2 = −1 i3 = −i i4 = 1 (a + bi)(a − bi) = a2 + b2 (a + bi) + (c + di) = a + c + (b + d)i (a + bi) − (c + di) = a − c + (b − d)i (a + bi)(c + di) = ac − bd + (ad + bc)i |a + bi| = √ a2 + b2 Complex Modulus (a + bi) = a − bi Complex Conjugate (a + bi)(a + bi) = |a + bi|2 DeMoivre’s Theorem Let z = r(cos θ + i sin θ), and let n be a positive integer. Then: Example: Let z = 1 − i, find z6. zn = rn(cos nθ + i sin nθ). Solution: First write z in polar form. r = √ (1)2 + (−1)2 = √ 2 θ = arg(z) = tan−1 −1 = − π Polar Form: z = √ 2 cos − π + i sin − π 4 4 Applying DeMoivre’s Theorem gives : z6 = √ 2 6 cos 6 · − π + i sin 6 · − π 4 4 = 23 cos − 3π + i sin − 3π 2 2 = 8(0 + i(1)) = 8i − − − − 10 More Conic Sections Hyperbola Standard Form for Horizontal Transverse Axis : (x h)2 a2 − (y k)2 b2 = 1 Standard Form for V ertical Transverse Axis : (y k)2 a2 − (x h)2 b2 = 1 Where (h, k)= center a=distance between center and either vertex Foci can be found by using b2 = c2 − a2 Where c is the distance between center and either focus. (b > 0) Parabola Vertical axis: y = a(x − h)2 + k Horizontal axis: x = a(y − k)2 + h Where (h, k)= vertex a=scaling factor f (x) = sin(x) 1 √3 2 √2 2 1 2 0 πππ 643 π 2 2π3π5π 3 4 6 π 7π5π4π 64 3 3π 2 5π7π 11π 3 4 6 2π −1 2 — 2 √2 — 2 -1 √3 Example : sin   5π √ 4 = − 2 2 f (x) = cos(x) 1 √3 2 √2 2 1 2 0 πππ 643 π 2 2π3π5π 346 π 7π5π4π 643 3π 2 5π7π 11π 346 2π −1 2 — 2 √2 — 2 -1 √3 Example : cos   7π √ 6 = − 3 2 11 f (x) x f (x) x π π f (x) = tan x √3 1 √3 3 −π — − −5π3π2π πππ 0 ππ 64 π 3 2π3π5π 643 — − −346 π — 3 √ 346 3 −1 −√3 12 − 2 f (x) 2 x