Section 7.7 Product-to-Sum and Sum-to-Product Formulas
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1 Section 7.7 Product-to-Sum and Sum-to-Product Fmulas Objective 1: Express Products as Sums To derive the Product-to-Sum Fmulas will begin by writing down the difference and sum fmulas of the cosine function: A cos(α β = cos(αcos(β + sin(αsin(β B cos(α + β = cos(αcos(β sin(αsin(β Now, add fmulas A and B together and solve f cos(αcos(β: A cos(α β = cos(αcos(β + sin(αsin(β B + cos(α + β = + cos(αcos(β sin(αsin(β cos(α β + cos(α + β = cos(αcos(β cos(αcos(β = cos(α β + cos(α + β 1 cos(αcos(β = 1 [cos(α β + cos(α + β] Next, we will subtract fmulas A and B and solve f sin(αsin(β: A cos(α β = cos(αcos(β + sin(αsin(β B cos(α + β = cos(αcos(β + sin(αsin(β cos(α β cos(α + β = sin(αsin(β sin(αsin(β = cos(α β cos(α + β sin(αsin(β = 1 [cos(α β cos(α + β] Finally, we will write down the difference and sum fmulas of the sine functions, add the fmulas together and solve f sin(αcos(β. C sin(α β = sin(αcos(β cos(αsin(β D + sin(α + β = + sin(αcos(β + cos(αsin(β sin(α β + sin(α + β = sin(αcos(β sin(αcos(β = sin(α β + sin(α + β 3 sin(αcos(β = 1 [sin(α β + sin(α + β] Product-to-Sum Fmulas 1 cos(αcos(β = 1 [cos(α β + cos(α + β] sin(αsin(β = 1 [cos(α β cos(α + β] 3 sin(αcos(β = 1 [sin(α β + sin(α + β]
2 11 Express as a sum difference containing only sines cosines. Ex. 1a cos(6θcos(θ Ex. 1b sin(3θsin(θ Ex. 1c sin(4θcos(3θ a cos(6θcos(θ = 1 [cos(6θ θ + cos(6θ + θ] = 1 cos(4θ + 1 cos(8θ b sin(3θsin(θ = 1 [cos(3θ θ cos(3θ + θ] = 1 cos(θ 1 cos(θ c sin(4θcos(3θ = 1 [sin(4θ 3θ + sin(4θ + 3θ] = 1 sin(θ + 1 sin(7θ Objective : Express Sums as Products. Now, we will go in the other direction to derive the Sum-to-Product Fmulas. We will first state the fmulas, and then prove them. Sum-to-Product Fmulas 4 sin(α + sin(β = sin( cos( sin(α sin(β = sin ( cos ( 6 cos(α + cos(β = cos ( cos ( 7 cos(α cos(β = sin ( sin ( Proof: 4 sin( cos( = ( 1 [sin ( + sin ( = ( 1 [sin β ( + sin ( α ] = sin(β + sin(α = sin(α + sin(β (apply product-to-sum #3 + ]
3 sin ( cos ( (apply product-to-sum #3 = ( 1 [sin ( + sin ( + ] = ( 1 [sin ( β α + sin( ] = sin( β + sin(α = sin(α + sin( β (sine function is odd = sin(α sin(β 6 cos ( cos ( (apply product-to-sum #1 = ( 1 [cos ( + cos ( = ( 1 [cos β α ( + cos( ] = cos(β + cos(α = cos(α + cos(β 7 sin( sin( = ( 1 [cos ( cos ( = ( 1 [cos β ( cos ( α ] = cos(β + cos(α = cos(α cos(β + ] (apply product-to-sum # + ] Express as a product containing only sines cosines. Ex. a sin(θ sin(θ Ex. b cos(6θ cos(4θ a sin(θ sin(θ (apply sum-to-product # = sin ( θ θ cos ( θ+θ = sin(θcos(3θ b cos(6θ cos(4θ (apply sum-to-product #7 = sin ( 6θ+4θ sin ( 6θ 4θ = sin(θsin(θ Establish the following identity: Ex. 3a Ex. 3b Ex. 3c sin(4θ+sin(θ cos(4θ+cos(θ = tan(3θ cos(α+cos(β = cot( cot( cos(α cos(β 1 cos(θ + cos(4θ cos(6θ = 4sin(θcos(θsin(3θ 1
4 13 a We will start on the left side: sin(4θ+sin(θ cos(4θ+cos(θ = sin( 4θ+θ cos( 4θ+θ = sin(3θcos(θ cos(3θcos(θ = sin(3θ cos(3θ = tan(3θ cos( 4θ θ cos( 4θ θ (use sum-to-product fmulas #4 & #6 (reduce b We will start from the left side. cos(α+cos(β cos(α cos(β = cos( sin( = cos( sin( = cot( cos( (simplify (use sum-to-product fmulas #6 & #7 sin( (reduce cos( sin( cot( (identity has been established c 1 cos(θ + cos(4θ cos(6θ (cos(0 = 1 = cos(0 cos(θ + cos(4θ cos(6θ (group = [cos(0 cos(θ] + [cos(4θ cos(6θ] (apply #7 = sin ( 0+θ 0 θ 4θ+6θ sin( sin( sin( 4θ 6θ (simplify = sin(θsin( θ sin(θsin( θ (sine is an odd function = sin(θsin(θ + sin(θsin(θ (fact out sin(θ = sin(θ[sin(θ + sin(θ] (apply #4 θ+θ θ θ ] (simplify = sin(θ[sin(3θcos( θ] (cosine is an even function = sin(θ[sin(3θcos(θ] (simplify = 4sin(θcos(θsin(3θ Identity is established. = sin(θ[sin ( cos (
5 Solve the following f angles in [0, π. Ex. 4 sin(4θ sin(6θ = 0 sin(4θ sin(6θ = 0 (sin(α sin(β = sin ( 4θ 6θ 4θ+6θ sin( cos( = 0 sin( θcos(θ = 0 sin(θcos(θ = 0 sin(θ = 0 cos(θ = 0 (sine function is odd cos ( θ = 0 π We will need to write down the general solution f θ: + πk + πk (divide by + πk + πk k = 0 k = 1 k = k = 3 k = 4 + π(0 + π(1 + π( + π(3 + π(4 = π = π = π = 9π = 1 = 17π + π(0 = + π(1 + π( + π(3 + π(4 = 11π = 7π = 1π = = 19π {0, π,, π, 7π, 9π 11π, π,, 1,, 17π, 19π } is the solution. Here is a listing of the imptant identities and fmulas from this chapter: 14 Fundamental Identities tan(θ = sin(θ cos(θ cot(θ = cos(θ sin(θ csc(θ = 1 sin(θ sec(θ = 1 cos(θ cot(θ = 1 tan(θ sin (θ + cos (θ = 1 tan (θ + 1 = sec (θ cot (θ + 1 = csc (θ sin( θ = sin(θ cos( θ = cos(θ tan( θ = tan(θ csc( θ = csc(θ sec( θ = sec(θ cot( θ = cot(θ
6 1 Sum and Difference Fmulas cos(α + β = cos(αcos(β sin(αsin(β cos(α β = cos(αcos(β + sin(αsin(β sin(α + β = sin(αcos(β + cos(αsin(β sin(α β = sin(αcos(β cos(αsin(β tan(α + β = tan(α+tan(β 1 tan(αtan(β tan(α β= tan(α tan(β 1+ tan(αtan(β Double-Angle Fmulas cos(θ = cos (θ sin (θ 1 sin (θ cos (θ 1 sin(θ = sin(θcos(θ Half angle fmulas sin( α = ± tan( α = ± tan( α = 1 cos(α sin(α 1 cos(α 1 cos(α 1+cos(α cos( α = ± tan(θ = tan(θ 1 tan (θ tan( α = sin(α 1+cos(α 1+cos(α Product-to-Sum Fmulas cos(αcos(β = 1 [cos(α β + cos(α + β] sin(αsin(β = 1 [cos(α β cos(α + β] sin(αcos(β = 1 [sin(α β + sin(α + β] Sum-to-Product Fmulas sin(α + sin(β = sin ( cos ( sin(α sin(β = sin ( cos ( cos(α + cos(β = cos( cos( cos(α cos(β = sin( sin(
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