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9.2 SUM AND DIFFERENCE FORMULAS FOR SINE AND COSINE Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally Sums and Differences of Angles Sum and Difference Formulas sin(θ + φ) = sin θ cos φ + cos θ sin φ sin(θ − φ) = sin θ cos φ − cos θ sin φ cos(θ + φ) = cos θ cos φ − sin θ sin φ cos(θ − φ) = cos θ cos φ + sin θ sin φ Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally Sums and Differences of Angles Example 1 Find an exact value for cos 15◦. Solution Let θ = 45◦ and φ = 15◦. Then cos 15◦ = cos(45◦ - 15◦) = cos 45◦ cos 15◦ + sin 45◦sin 15◦ = 2 3 2 1 2 2 6 2 2 2 4 We can check our answer using a calculator: 6 4 2 0 . 9659 and cos 15◦ = 0.9659. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally Justification of cos(θ − φ) = cos θ cos φ + sin θ sin φ y To derive the formula for cos(θ − φ), we find the ●A 1 distance between points A and B in the figure in θ ●B φ x two different ways. These points correspond to the O angles θ and φ on the unit circle, so their coordinates are A = (cosθ, sinθ) and B = (cos φ, sin φ). The angle AOB is (θ − φ), so if we use the Law of Cosines, the distance AB is given by AB2 = 12 + 12 − 2 · 1 · 1 cos(θ − φ) = 2 − 2 cos(θ − φ). Next, we find the distance between A and B using the distance formula and multiply out AB2 = (cos θ − cos φ)2 + (sin θ − sin φ)2 distance formula = cos2θ − 2cosθ cosφ + cos2φ + sin2θ − 2sinθ sinφ + sin2 φ = cos2θ + sin2θ + cos2φ + sin2φ − 2cosθ cosφ − 2sinθ sin φ = 2 − 2(cos θ cos φ + sin θ sin φ). Setting the two expressions for the distance equal gives what we wanted: 2 − 2 cos(θ − φ) = 2 − 2(cos θ cos φ + sin θ sin φ) so cos(θ − φ) = cos θ cos φ + sin θ sin φ. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally Sums and Differences of Sines and Cosines: Same Amplitudes What can we say about f(t) = sin t + cos t? y = 2 y = sin t + cos t The graph looks remarkably like the graph of the sine function, having the same period but a larger amplitude (somewhere between 1 and 2). The horizontal shift appears to be about 1/8 period, so the phase shift is φ = π /4. What about the value of A? Assuming φ = π/4, let’s use our identity for sin(θ + φ) to rewrite sin(t + π/4). Substituting θ = t and φ = π/4, we obtain sin(t + π/4) = cos π/4 sin t + sin π/4 cos t = (1/ 2 ) (sin t + cos t) or sin t + cos t = 2 sin(t + π/4) Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally Rewriting a1 sinBt + a2 cosBt We start with a1 sinBt + a2 cosBt, where a1, a2, and B are constants. We want to write this expression in the form A sin(Bt + φ). To do this, we show that we can find the constants A and φ for any a1 and a2. We use the identity sin(θ + φ) = sin θ cos φ + sin φ cos θ with θ = Bt: sin(Bt + φ) = sin(Bt) cos φ + sin φ cos(Bt) Multiplying by A, we have A sin(Bt + φ) = A cos φ sin(Bt) + A sin φ cos(Bt). Letting a1 = A cos φ and a2 = A sin φ ,we have A sin(Bt + φ) = a1 sin(Bt) + a2 cos(Bt). We want to write A and φ in terms of a1 and a2; so far, we know a1 and a2 in terms of A and φ. Since 2 2 a12+ a22 = A2 cos2 φ + A2 sin2 φ = A2, then A a 1 a 2 . Also, a2 / a1 = (A sin φ)/(A cos φ) = sin φ/cos φ = tan φ. These formulas allow A and φ to be determined from a1 and a2. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally Rewriting a1 sinBt + a2 cosBt Example 4 Check algebraically that sin t + cos t = 2 sin(t + π/4). Solution Here, a1 = 1 and a2 = 1, so A = 12 12 2 . And tan φ = a2/a1 = 1 / 1 = 1, So φ = tan-1 1 = π/4 This confirms what we already knew, namely that sin t + cos t = 2 sin(t + π/4). Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally Rewriting a1 sinBt + a2 cosBt Provided their periods are equal, the sum of a sine function and a cosine function can be written as a single sinusoidal function. We have a1 sin(Bt) + a2 cos(Bt)= A sin(Bt + φ) 2 2 where A a 1 a 2 and φ = tan-1 (a2 / a1). The angle φ is determined by the equations cos φ = a1 /A and sin φ = a2 /A. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally Sums and Differences of Sines and Cosines: Same Amplitudes We begin with the fact that cos(θ + φ) = cos θ cos φ − sin θ sin φ cos(θ − φ) = cos θ cos φ + sin θ sin φ. If we add these two equations, the terms involving the sine function cancel out, giving cos(θ + φ) + cos(θ − φ) = 2 cos θ cos φ . The left side of this equation can be rewritten by substituting u = θ + φ and v = θ − φ : cos u + cos v = 2 cos θ cos φ . The right side of the equation can be rewritten in terms of u and v: u + v = (θ + φ) + (θ − φ) = 2θ and u – v = (θ + φ) – (θ − φ) = 2φ. So Θ = (u + v)/2 and φ = (u – v )/2 and the above equation becomes cos u + cos v = 2 cos((u + v)/2) cos((u − v)/2) Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally Sums and Differences of Sines and Cosines: Same Amplitudes Example 6 Write cos(30t) + cos(28t) as the product of two cosine functions. Solution Letting u = 30t, v = 28t , we have cos(30t) + cos(28t) = 2 cos((30t + 28t)/2) · cos((30t − 28t)/2) = 2 cos(29t) · cos t. Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally Sums and Differences of Sines and Cosines: Same Amplitudes Sum and Difference of Sine and Cosine cos u + cos v = 2 cos((u + v)/2) cos((u − v)/2) sin u + sin v = 2 sin((u + v)/2) cos((u − v)/2) cos u − cos v = 2 sin((u + v)/2) sin((u − v)/2) sin u − sin v = 2 cos((u + v)/2) sin((u − v)/2) Functions Modeling Change: A Preparation for Calculus, 4th Edition, 2011, Connally