1. IDENTITIES, EXPRESSIONS, AND EQUATIONS
9.1
IDENTITIES, EXPRESSIONS, AND EQUATIONS
Functions Modeling Change: A Preparation for Calculus, th Edition, 2011, Connally

2. Deriving New Identities
Example 1 The Pythagorean identity, cos2 θ + sin2 θ = 1, can be rewritten in terms of other trigonometric functions. Provided cos θ ≠ 0, dividing through by cos2 θ gives The identity we derived is another version of the Pythagorean identity. It is usually written
1 + tan2 θ = sec2 θ.
Functions Modeling Change: A Preparation for Calculus, th Edition, 2011, Connally

3. 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 = cos t 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 (cos2 t + sin2 t) r = cos t, 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, th Edition, 2011, Connally

4. Using Identities to Evaluate Expressions
Example 3 Suppose that cos θ = 2/3 and 3π/2 ≤ θ ≤ 2π. Find sin θ and tan θ Solution Use the relationship cos2 θ + sin2 θ = 1 to find sin θ. Substitute cos θ = 2/3: (2/3)2 + sin2 θ = 1 4/9 + sin2 θ = 1 sin2 θ = 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 / cos θ = (− )/(2/3) = −
Functions Modeling Change: A Preparation for Calculus, th Edition, 2011, Connally

5. Using Identities to Solve Equations
Example 6 Solve 2 sin2 t = 3 − 3 cos t for 0 ≤ t ≤ π. Solution We can use the identity sin2 t = 1 − cos2 t to rewrite this equation entirely in terms of cos t: 2 sin2 t = 2(1 − cos2 t) = 2− 2cos2 t = 3 − 3 cos t 2 cos2 t − 3 cos t + 1 = 0. It can be helpful to abbreviate cos t = x, so x2 = (cos t)2 = cos2 t, giving: 2x2 − 3x + 1 = 0 (2x − 1)(x − 1) = 0 so x = ½ or x = 1. Since x = cos t, this means either cos t = ½ with solution t = π /3, or cos t = 1 with solution t = 0.
Functions Modeling Change: A Preparation for Calculus, th Edition, 2011, Connally

6. Double-Angle Formula for Sine
Thinking algebraically, we would like to find a formula for sin 2θ in terms of sin θ and cos θ. 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 cos θ, and the hypotenuse is 1, so
sin α = Opposite / Hypotenuse = cos θ / 1 = cos θ
Thus, substituting cos θ for sin α, we have
(sin 2θ)/(2 sin θ) = cos θ A 1 α
sin θ θ B O θ
cos θ
sin θ 1 C
sin 2θ = 2 sin θ cos θ
Functions Modeling Change: A Preparation for Calculus, th Edition, 2011, Connally

7. 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 t cos t = 2 sin t 2 sin t cos t − 2 sin t = 0 2 sin t (cos t − 1) = 0 factoring out 2 sin t. Thus, 2 sin t = 0 or cos t = 1. Now, 2 sin t = 0 for t = 0, π, and 2 π, and cos t = 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, th Edition, 2011, Connally

8. 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 2t and y = 2 sin t intersect.
y = 2 sin t
y = sin 2t
Functions Modeling Change: A Preparation for Calculus, th Edition, 2011, Connally

9. 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, th Edition, 2011, Connally

10. SUM AND DIFFERENCE FORMULAS FOR SINE AND COSINE
9.2
SUM AND DIFFERENCE FORMULAS FOR
SINE AND COSINE
Functions Modeling Change: A Preparation for Calculus, th Edition, 2011, Connally

11. 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, th Edition, 2011, Connally

12. 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◦ = We can check our answer using a calculator: and cos 15◦ =
Functions Modeling Change: A Preparation for Calculus, th Edition, 2011, Connally

13. Justification of cos(θ − φ) = cos θ cos φ + sin θ sin φy
To derive the formula for cos(θ − φ), we find the distance between points A and B in the figure in two different ways. These points correspond to the 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 = − 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. ●A 1 θ ●B φ x O
Functions Modeling Change: A Preparation for Calculus, th Edition, 2011, Connally

14. Sums and Differences of Sines and Cosines: Same Amplitudes
What can we say about f(t) = 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/ ) (sin t + cos t) or sin t + cos t = sin(t + π/4)
y =
y = sin t + cos t
Functions Modeling Change: A Preparation for Calculus, th Edition, 2011, Connally

15. 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 a12+ a22 = A2 cos2 φ + A2 sin2 φ = A2, then . 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, th Edition, 2011, Connally

16. Rewriting a1 sinBt + a2 cosBt
Example 4 Check algebraically that sin t + cos t = sin(t + π/4). Solution Here, a1 = 1 and a2 = 1, so A = . 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 = sin(t + π/4).
Functions Modeling Change: A Preparation for Calculus, th Edition, 2011, Connally

17. 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) where 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, th Edition, 2011, Connally

18. 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, th Edition, 2011, Connally

19. 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, th Edition, 2011, Connally

20. 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, th Edition, 2011, Connally

21. 9.3
TRIGONOMETRIC MODELS
Functions Modeling Change: A Preparation for Calculus, th Edition, 2011, Connally

22. Sums of Trigonometric Functions
Example 2 Sketch and describe the graph of y = sin 2x + sin 3x. Solution The figure shows that the function y = sin 2x + sin 3x is not sinusoidal. It is, however, periodic. Its period seems to be 2π, since it repeats twice on the interval of length 4 π shown in the figure.
Since the period of sin 2x is π and the period of sin 3x is 2π/3, on any interval of length 2π, y = sin 2x completes two cycles and y = sin 3x completes three cycles. Both are at the beginning of a new cycle after an interval of 2π, so their sum begins to repeat at this point. Even though the maximum value of each is 1, the maximum value of their sum is not 2; it is a little less than 2, because they achieve their maximum values for different x values.
y=2
y = sin 2x + sin 3x
y=sin 3x
y=sin 2x
Functions Modeling Change: A Preparation for Calculus, th Edition, 2011, Connally

23. Damped Oscillation
If A0, B, and C, and k are constants, k > 0, a function of the form y = A0e−kt cos(Bt) + C or y = A0e−kt sin(Bt) + C can be used to model an oscillating quantity whose amplitude decreases exponentially according to A(t) = A0e−kt where A0 is the initial amplitude. Our model for the displacement of a weight is in this form with k = ln 2.
A graph of the weight’s displacement
assuming that the amplitude is a decreasing
exponential function of time
The weight is still making small oscillations for t ≥ 5. Note reduced scale on d-axis; t-axis starts at t = 5
Functions Modeling Change: A Preparation for Calculus, th Edition, 2011, Connally

24. Oscillation With a Rising Midline
Example 3 In Section 8.2, Example 6, we represented a rabbit population undergoing seasonal fluctuations by the function R = f(t) = − 5000 cos(π/6 t), where R is the size of the rabbit population t months after January. Now let us imagine a different situation. What if the average, even over long periods of time, does not remain constant? For example, suppose that, due to conservation efforts, there is a steady increase of 50 rabbits per month in the average rabbit population. Thus, we could write R = 10, t − 5000 cos(π/6 t)
R = 10, t
A graph of the gradually increasing rabbit population
Functions Modeling Change: A Preparation for Calculus, th Edition, 2011, Connally

25. Acoustic Beats
Suppose a frequency of 55 hertz is struck on a tuning fork together with a note on an out-of-tune piano, whose frequency is 61 hertz. The intensities, I1 and I2, of these two tones are represented by the functions I1 = cos(2πf1t) and I2 = cos(2πf2t), where f1 = 55, and f2 = 61, and t is in seconds. If both tones are sounded at the same time, then their combined intensity is the sum of their separate intensities: I = I1 + I2 = cos(2πf1t) + cos(2πf2t). The graph of this function in the figure on the next slide resembles a rapidly varying sinusoidal function except that its amplitude increases and decreases. The ear perceives this variation in amplitude as a variation in loudness, so the tone appears to waver (or beat) in a regular way. This is an example of acoustic beats. A piano can be tuned by adjusting it until the beats fade.
Functions Modeling Change: A Preparation for Calculus, th Edition, 2011, Connally

26. Acoustic Beats
Graph of I = cos(2π ˑ55 t) + cos(2πˑ 61 t) How can we explain this graph? Using the identity for a sum of cosines we rewrite the intensity as I = cos(2π 61 t) + cos(2π 55 t) = 2 cos(2π (61+55)/2 t) cos(2π (61-55)/2 t) = 2 cos(2π · 58 t) cos(2π · 3 t) We can think of this formula in the following way: I = = 2 cos(2π · 58 t) cos(2π · 3 t) = A(t) p(t) where A(t) = 2 cos(2π ˑ58 t) gives a (slowly) changing amplitude and p(t) = cos(2π · 3 t) gives a pure tone of 58 hz.
Functions Modeling Change: A Preparation for Calculus, th Edition, 2011, Connally

27. Acoustic Beats
The function A(t) = 2 cos(2π · 3t) gives a varying amplitude, while the function p(t) = cos(2π · 58t) gives a pure tone of 58 hz.
Thus, we can think of the tone described by I as having a pitch of 58 hz,
which is midway between the tones sounded by the tuning fork and the out-of-tune piano. As the amplitude rises and falls, the tone grows louder and softer, but its pitch remains a constant 58 hz. Notice from the figure that the function A(t) completes three full cycles on the interval 0 ≤ t ≤ 1. The tone is loudest when A(t) = 1 or A(t) = −1. Since both of these values occur once per cycle, the tone grows loud six times every second.
Tone is loud when |A(t)| is a maximum.
A(t) gives varying amplitude
p(t) is a pure tone of 58 hz.
Tone is soft when A(t) is zero.
Functions Modeling Change: A Preparation for Calculus, th Edition, 2011, Connally