9.1 IDENTITIES, EXPRESSIONS, AND EQUATIONS Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally
Deriving New Identities Example 1 The Pythagorean identity, cos2θ + sin2θ = 1, can be rewritten in terms of other trigonometric functions. Provided cosθ≠ 0, dividing through by cos2θ gives The identity we derived is another version of the Pythagorean identity. It is usually written 1 + tan2θ= sec2θ. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally
Using Identities To Simplify Expressions Example 3 Simplify the expression (2 cos t + 3 sin t) (3 cos t + 2 sin t) − 13 sin t cos t. Solution To make the calculations easier, let r = cost and s = sin t. Our expression becomes (2r + 3s)(3r + 2s) − 13rs = 6r2 + 4rs + 9rs + 6s2 − 13rs multiply out = 6r2 + 6s2 + 13rs − 13rs regroup = 6 (r2 + s2) simplify and factor = 6 (cos2t + sin2t) r = cost, s = sin t = 6. because cos2 t + sin2 t = 1 We see that this complicated expression equals 6. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally
Using Identities to Evaluate Expressions Example 3 Suppose that cosθ = 2/3 and 3π/2 ≤ θ ≤ 2π. Find sin θ and tan θSolution Use the relationship cos2θ + sin2θ = 1 to find sin θ. Substitute cosθ = 2/3: (2/3)2 + sin2θ = 1 4/9 + sin2θ = 1 sin2θ = 1 − 4/9 = 5/9 sin θ = Because θ is in the fourth quadrant, sin θ is negative, so sin θ = − . To find tan θ, use the relationship tan θ = sin θ / cosθ = (− )/(2/3) = − Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally
Using Identities to Solve Equations Example 6 Solve 2 sin2t = 3 − 3 cos t for 0 ≤ t ≤ π. Solution We can use the identity sin2t = 1 − cos2t to rewrite this equation entirely in terms of cost: 2 sin2t = 2(1 − cos2t) = 2− 2cos2t = 3 − 3 cos t 2 cos2t − 3 cost + 1 = 0. It can be helpful to abbreviate cost = x, so x2 = (cost)2 = cos2t, giving: 2x2 − 3x + 1 = 0 (2x − 1)(x − 1) = 0 so x = ½ or x = 1. Since x = cost, this means either cost = ½ with solution t = π /3, or cost = 1 with solution t = 0. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally
Double-Angle Formula for Sine Thinking algebraically, we would like to find a formula for sin 2θ in terms of sin θ and cosθ. We derive our formula by using the figure. The lengths of OA and OC are 1; the length of AC is 2sin θ. Writing for the angle at A and applying the Law of Sines to triangle OAC gives (sin 2θ) / (2 sin θ) = sin α / 1 In triangle OAB, the length of side OB is cosθ, and the hypotenuse is 1, so sin α = Opposite / Hypotenuse = cosθ / 1 = cosθ Thus, substituting cosθ for sin α, we have (sin 2θ)/(2 sin θ) = cosθ sin 2θ= 2 sin θcos θ A α 1 sin θ θ B θ cosθ O sin θ 1 C Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally
Using Identities to Solve Equations Example 7 Find all solutions to the equation sin 2t = 2 sin t on the interval 0 ≤ t ≤ 2π. Solution Using the double-angle formula sin 2t = 2 sin t cos t, we have 2 sin tcost = 2 sin t 2 sin tcost − 2 sin t = 0 2 sin t (cost − 1) = 0 factoring out 2 sin t. Thus, 2 sin t = 0 or cost = 1. Now, 2 sin t = 0 for t = 0, π, and 2 π, and cost = 1 for t = 0 and 2 π. Thus there are three solutions to the original equation on the interval 0 ≤ t ≤ 2 π : t = 0, π, and 2 π. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally
Using Identities to Solve Equations Example 7 (continued) Find all solutions to the equation sin 2t = 2 sin t on the interval 0 ≤ t ≤ 2π. Solution The figure illustrates the solutions (t = 0, π, and 2 π) graphically as the points where the graphs of y = sin 2tand y = 2 sin tintersect. y = 2 sin t y = sin 2t Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally
Double-Angle Formulas for Cosine and Tangent The double-angle formula for the cosine can be written in three forms: cos 2θ= 1 − 2 sin2θ cos 2θ= 2 cos2θ− 1 cos 2θ= cos2θ− sin2θ tan 2θ = (2 tan θ) / (1 − tan2 θ) Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally
