WARM UP What is the exact value of cos 30°? What is the exact value of tan 60°? Write cos 57° in decimal form. Write sin 33° in decimal form. Write csc 9° in decimal form.

SINUSOID COMBINATIONS COMPOSITE ARGUMENT & LINEAR COMBINATION PROPERTIES

OBJECTIVES Derive a composite argument property expressing cos (A – B) in terms of cosines and sines of A and B. Express a linear combination of cosine and sine with the same period as a single cosine with a phase displacement.

KEY TERMS & CONCEPTS In phase Linear combination Composite argument property Distance formula Complement

INTRODUCTION If you add two sinusoids such as and, you get another sinusoid with amplitude equal to the sum of the two amplitudes. These sinusoids are said to be in phase because their high and low points occur at the same values of θ.

INTRODUCTION If you add two sinusoids with the same period that are not in phase, the result is also a sinusoid, with amplitude less than the sum you found in section 5-1. We will now be studying algebraic ways to find the amplitude and phase displacement of a linear combination of cosine and sine, in the form of y = a cos θ + b sin θ.

INTRODUCTION With the help of the composite argument property by which you can express cos (A – B) in terms of cosines and sines of A and B, you will prove that any linear combination of cosine and sine with equal periods is another sinusoid with the same period. This principle governs the sound produced by musical instruments. Two instruments playing middle C produce the sound of middle C whether or not the sound waves are in phase.

LINEAR COMBINATION PROPERTY The graph of y = 3 cos θ + 4 sin θ is a sinusoid that you can express in the form y = A cos (θ – D) where A is the amplitude and D is the phase displacement for y = cos x. If you plot the graph and use the maximum feature on your grapher, you will find A = 5 and D = …° D is the angle in standard position with u = 3 (coefficient of cosine) and v = 4 (coefficient of sine). After deriving the composite argument property, you can prove that A is the length of the hypotenuse of the reference triangle for D.

LINEAR COMBINATION PROPERTY You can find A using the Pythagorean theorem and D by applying the concept of arctangent. D = arctan = …° + 180n° = …° Choose n = 0 so that D terminates in Quadrant I. So y = 5 cos (θ – …°) is equivalent to y = 3 cos θ + 4 sin θ. The graphical solution confirms this.

EXAMPLE 1 Express y = − 8 cos θ + 3 sin θ as a single cosine with a phase displacement. Solution: Sketch angle D in standard position with u = -8 and v = 3 D = arctan 3/-8 = − …° + 180n° Find A by the Pythagorean theorem Find D using the definition of arctangent = …° Choose n = 1 to place D in the correct quadrant.

PROPERTY Linear Combination and Sine with Equal Periods b cos x + c sine x = A cos (x – D) where and D = arctan The quadrant for D = arctangent depends on the signs of b and c and may be determined by sketching D in standard position. Then length of the hypotenuse of the reference triangle is A.

WARM UP Write the liner combination of cosine & sine as a single cosine with a phase displacement. 1. y = − 8 cos θ + 3 sin θ 2. y = 4 cos θ + 3 sin θ 3. y = -7 cos θ + 24 sin θ 4. y = -15 cos θ + 8 sin θ 13 cos(θ – ) 5 cos(θ – ) 25 cos(θ – ) 17 cos(θ – )

COMPOSITE ARGUMENT PROPERTY FOR COSINE (A – B) You will recall that the cosine function does not distribute over addition or subtraction. Consider this proof by counterexample.

COMPOSITE ARGUMENT However, you can express cos (58° - 20°) exactly in terms of sines and cosines of 58° and 20°. The result is Both sides equal …

CONTINUED Next you’ll see how to generalize the results for any angles A and B. The left side of the figure shows angles A and B in standard position and shows their difference, angle (A – B). The coordinates of the points where the initial and terminal sides of angle (A – B) cut the unit circle are cos A, sin A and cos B, sin B

DISTANCE FORMULA The chord between the initial and terminal points has length d, which can be calculated by the Pythagorean theorem. As shown on the right side of the figure, d is the hypotenuse of a right triangle with side lengths (cos A – cos A) and sin A – sin A). You may recall this as the distance formula. By the Pythagorean theorem Expand the squares Commute and associate the squared terms 2 – 2 cos A cos B – 2 sin A sin B – 2 cos A cos B – 2 sin A sin B

CONTINUED The figure below shows angle (A – B) rotated into standard position. The coordinates of the terminal and initial points of the angle in this position are cos (A – B), sin (A – B)and (1, 0)

CONTINUED The chord still has length d, By the distance formula and subsequent algebra Expand the squares. = [cos 2 (A − B) + sin 2 (A − B)] + 1 − 2 cos (A − B) Associate the squared terms. d 2 = 2 − 2 cos (A − B) Use the Pythagorean property This is the property that was illustrated by numerical example with A = 58°and B = 20°.

PROPERTY Composite Argument Property for cos (A – B) cos (A – B) = cos A cos B + sin A sin B Verbally: Cosine of the difference of two angles is equal to cosine of first angle times cosine of second angle, plus sine of first anle times sine of second angle.

EXAMPLE 2 Express 7 cos (θ - 23°) as a linear combination of cos θ and sin θ. Solution: 7 cos (θ − 23°) = 7 (cos θ cos 23° + sin θ sin 23°) Apply the composite argument property = 7 cos θ cos 23° + 7 sin θ sin 23° = (7 cos 23°) cos θ + (7 sin 23°) sin θ Apply the composite argument property 7 cos (θ − 23°) = … cos θ … sin θ A linear combination of cos θ and sin θ

WARM UP Express each equation as a linear combination of cosine and sine. 1.y = 10 cos (θ – 30°) 2.y = 20 cos (θ – 60°) 3.y = 8 cos (2θ – 120°) 5√3 cos θ + 5 sin θ 10 cos θ + 10√3 sin θ -4 cos 2θ + 4√3 sin 2θ

ALGEBRAIC SOLUTION OF EQUATIONS You can use the linear combination property to solve certain equations algebraically. Example 3: Solve -2 cos x + 3 sin x = 2 for x [-2π, 2π]. Verify the solution graphically.

SOLUTION -2 cos x + 3 sin x = 2 Transform the left side, -2 cos x + 3 sin x, into the form A cos (x – D). Draw angle D in standard position Write the given equation Use the Pythagorean theorem to calculate A. Use n = 1 for the proper arctangent value. Rewrite the equation using A cos (x – D).

SOLUTION CONTINUED Evaluate … …and … – Pick values of n to get x in the domain. or

SOLUTION CONTINUED This the graph of y = -2 cos x + 3 sin x and the line y = 2. Note that the graph is a sinusoid, as you discovered algebraically while solving the equation. By using the intersect feature, you can see that the four solutions are correct and that they are the only solution in the domain [-2π, 2π]. The graph also show the phase displacement …and the amplitude or approximately 3.6

CH 5.1/5.2 HOMEWORK Textbook pg. 197 #1 – 7 & Journal Question #8 Pg 203 #1 – 4 Pg #6, 10, 14, 20, 22, and Journal Question #33.