### Section 9.2

```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
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