Trigonometric Formula Sheet

Trigonometric Formula Sheet Definition of the Trig Functions Unit Circle Definition Assume θ can be any angle. Right Triangle Definition Assume that: 0 < θ < π2 or 0◦ < θ < 90◦ y (x, y) 1 hypotenuse opposite y θ x x θ adjacent opp hyp adj cos θ = hyp opp tan θ = adj sin θ = hyp cosec θ = opp hyp sec θ = adj adj cot θ = opp y 1 x cos θ = 1 y tan θ = x sin θ = 1 y 1 sec θ = x x cot θ = y csc θ = Domains of the Trig Functions sin θ, ∀ θ ∈ (−∞, ∞) cosec θ, ∀ θ 6= nπ, where n ∈ Z cos θ, ∀ θ ∈ (−∞, ∞) sec θ, tan θ, 1 π, where n ∈ Z ∀ θ 6= n + 2  cot θ,  1 ∀ θ 6= n + π, where n ∈ Z 2 ∀ θ 6= 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. 2π ω 2π cos(ωθ) ⇒ T = ω π tan(ωθ) ⇒ T = ω sin(ωθ) ⇒ T = 2π ω 2π sec(ωθ) ⇒ T = ω π cot(ωθ) ⇒ T = ω csc(ωθ) ⇒ T = Identities and Formulas Tangent and Cotangent Identities tan θ = sin θ cos θ cot θ = cos θ sin θ Reciprocal Identities 1 csc θ 1 cos θ = sec θ 1 tan θ = cot θ sin θ = 1 sin θ 1 sec θ = cos θ 1 cot θ = tan θ csc θ = Pythagorean Identities 2 2 sin θ + cos θ = 1 tan2 θ + 1 = sec2 θ csc(−θ) = − csc θ sec(−θ) = sec θ cot(−θ) = − cot θ Periodic Formulas If n is an integer csc(θ + 2πn) = csc θ sec(θ + 2πn) = sec θ cot(θ + πn) = cot θ Double Angle Formulas sin(2θ) = 2 sin θ cos θ cos(2θ) = cos2 θ − sin2 θ = 2 cos2 θ − 1 = 1 − 2 sin2 θ tan(2θ) = 2 tan θ 1 − tan2 θ sin(α ± β) = sin α cos β ± cos α sin β cos(α ± β) = cos α cos β ∓ sin α sin β tan α ± tan β 1 ∓ tan α tan β Product to Sum Formulas Even and Odd Formulas sin(θ + 2πn) = sin θ cos(θ + 2πn) = cos θ tan(θ + πn) = tan θ Sum and Difference Formulas tan(α ± β) = 1 + cot2 θ = csc2 θ sin(−θ) = − sin θ cos(−θ) = cos θ tan(−θ) = − tan θ Half Angle Formulas r 1 − cos(2θ) sin θ = ± 2 r 1 + cos(2θ) cos θ = ± 2 s 1 − cos(2θ) tan θ = ± 1 + cos(2θ) 1 sin α sin β = [cos(α − β) − cos(α + β)] 2 1 cos α cos β = [cos(α − β) + cos(α + β)] 2 1 sin α cos β = [sin(α + β) + sin(α − β)] 2 1 cos α sin β = [sin(α + β) − sin(α − β)] 2 Sum to Product Formulas     α+β α−β sin α + sin β = 2 sin cos 2 2     α+β α−β sin α − sin β = 2 cos sin 2 2     α+β α−β cos α + cos β = 2 cos cos 2 2     α+β α−β cos α − cos β = −2 sin sin 2 2 Cofunction Formulas π  π  Degrees to Radians Formulas sin − θ = cos θ cos − θ = sin θ 2 2 If x is an angle in degrees and t is an angle in π  π  radians then: csc − θ = sec θ sec − θ = csc θ 2 2 π  π  π t πx 180◦ t = ⇒ t = and x = tan − θ = cot θ cot − θ = tan θ 180◦ x 180◦ π 2 2 Unit Circle (0, 1) (− 12 , (− 3 ) 2 ( 12 , √ 3 ) 2 √ √ 2 2 , ) 2 2 √ ( 60◦ , π3 120◦ , 2π 3 √ 2 2 , ) 2 2 ( √ 3 1 , ) 2 2 30◦ , π6 150◦ , 5π 6 (−1, 0) √ 45◦ , π4 135◦ , 3π 4 3 1 , ) 2 2 (− 90◦ , π2 √ 180◦ , π 0◦ , 2π 210◦ , 7π 6 (− √ 3 , − 12 ) 2 (− √ 330◦ , 11π 6 225◦ , 5π 4 240◦ , 4π 3 √ 2 , − 22 ) 2 300◦ , 5π 3 ( (− 12 , − √ 3 ) 2 270◦ , 3π 2 ( 12 , − √ (0, −1) F or any ordered pair on the unit circle (x, y) : cos θ = x and sin θ = y Example cos ( 7π 6) ( 315◦ , 7π 4 = √ − 23 1 sin ( 7π 6 ) = −2 3 ) 2 √ √ 3 , − 21 ) 2 √ 2 2 , − ) 2 2 (1, 0) Inverse Trig Functions Definition Inverse Properties These properties hold for x in the domain and θ in the range θ = sin−1 (x) is equivalent to x = sin θ θ = cos−1 (x) is equivalent to x = cos θ θ = tan−1 (x) is equivalent to x = tan θ Domain and Range Function Domain Range θ = sin−1 (x) −1 ≤ x ≤ 1 − −1 θ = cos (x) −1 θ = tan (x) −∞ ≤ x ≤ ∞ sin−1 (sin(θ)) = θ cos(cos−1 (x)) = x cos−1 (cos(θ)) = θ tan(tan−1 (x)) = x tan−1 (tan(θ)) = θ Other Notations π π ≤θ≤ 2 2 sin−1 (x) = arcsin(x) 0≤θ≤π −1 ≤ x ≤ 1 sin(sin−1 (x)) = x cos−1 (x) = arccos(x) π π − <θ< 2 2 tan−1 (x) = arctan(x) Law of Sines, Cosines, and Tangents β a c γ α b Law of Sines sin α sin β sin γ = = a b c Law of Cosines a2 = b2 + c2 − 2bc cos α b2 = a2 + c2 − 2ac cos β c2 = a2 + b2 − 2ab cos γ Law of Tangents tan 12 (α − β) a−b = a+b tan 12 (α + β) tan 12 (β − γ) b−c = b+c tan 12 (β + γ) tan 12 (α − γ) a−c = a+c tan 21 (α + γ) Finding the nth roots of a number using DeMoivre’s Theorem Example: Find all the complex fourth roots of 4. That is, find all the complex solutions of x4 = 4. We are asked to find all complex fourth roots of 4. These are all the solutions (including the complex values) of the equation x4 = 4. For any positive integer n , a nonzero complex number z has exactly n distinct nth roots. More specifically, if z is written in the trigonometric form r(cos θ + i sin θ), the nth roots of z are given by the following formula.      θ 360◦ k θ 360◦ k + + + i sin , f or k = 0, 1, 2, ..., n − 1. (∗) r cos n n n n 1 n Remember from the previous example we need to write 4 in trigonometric   form by using: p b r = (a)2 + (b)2 . and θ = arg(z) = tan−1 a So we have the complex number a + ib = 4 + i0. Therefore a = 4 and b = 0 p (4)2 + (0)2=  4 and 0 θ = arg(z) = tan−1 =0 4 Finally our trigonometric form is 4 = 4(cos 0◦ + i sin 0◦ ) So r = Using the formula (∗) above with n = 4, we can find the fourth roots of 4(cos 0◦ + i sin 0◦ )   ◦   ◦  √ √ 1 0 360◦ ∗ 0 360◦ ∗ 0 0 • For k = 0, 4 4 cos + + + i sin = 2 (cos(0◦ ) + i sin(0◦ )) = 2 4 4 4 4   ◦   ◦  ◦ ◦ √ √ 1 0 360 ∗ 1 0 360 ∗ 1 4 • For k = 1, 4 cos + + i sin + = 2 (cos(90◦ ) + i sin(90◦ )) = 2i 4 4 4 4   ◦   ◦  ◦ √ √ 1 0 360 ∗ 2 0 360◦ ∗ 2 • For k = 2, 4 4 cos + + i sin + = 2 (cos(180◦ ) + i sin(180◦ )) = − 2 4 4 4 4   ◦   ◦  ◦ √ √ 1 0 360 ∗ 3 0 360◦ ∗ 3 • For k = 3, 4 4 cos + + i sin + = 2 (cos(270◦ ) + i sin(270◦ )) = − 2i 4 4 4 4 Thus all of the complex roots of x4 = 4 are: √ √ √ √ 2, 2i, − 2, − 2i . f (x) f (x) = sin(x) 1√ 3 2 √ 2 2 1 2 x 0 π 6 π 4 π 3 π 2 2π 3 3π 4 5π 6 π 7π 6 5π 4 4π 3 3π 2 5π 3 7π 4 11π 6 2π 4π 3 3π 2 5π 3 7π 4 11π 6 2π − 12 √ 2 2 √ − 23 − -1 √  5π  2 Example : sin =− 4 2   f (x) f (x) = cos(x) 1√ 3 2 √ 2 2 1 2 x 0 π 6 π 4 π 3 π 2 2π 3 3π 4 5π 6 π 7π 6 5π 4 − 12 √ 2 2 √ − 23 − -1 √  7π  3 =− Example : cos 6 2   − π2 π 2 f (x) f (x) = tan x √ 3 1 √ 3 3 −π 3π 2π − 5π 6 − 4 − 3 − π3 − π4 − π6 0 − √ 3 3 −1 √ − 3 x π 6 π 4 π 3 2π 3 3π 4 5π 6 π