To hit the target 900 meters away, we want x( t) =900at the time when the cannonball
hits the ground, when h( t) = 0 . To solve this problem, we will first solve for the time, t, when the cannonball hits the ground. Our answer will depend upon the angleθ .
h( t) = 0
− 9
.
4 2
t +100sin(θ) t = 0
Factor
t(− 9
.
4 t +100sin(θ )) = 0
Break this apart to find two solutions
t =0 or
− .
4 9 t +100sin(θ ) = 0
Solve for t
− 9
.
4 t = 100
−
sin(θ )
100sin(θ )
t =
9
.
4
This shows that the height is 0 twice, once at t = 0 when the cannonball is fired, and
again when the cannonball hits the ground after flying through the air. This second
value of t gives the time when the ball hits the ground in terms of the angle θ . We
want the horizontal distance x(t) to be 900 when the ball hits the ground, in other words
θ
when
100sin( )
t =
.
9
.
4
Since the target is 900 m away we start with
x( t) =900
Use the formula for x(t)
100cos(θ) t =900
Substitute the desired time, t from above
100sin(θ )
100cos(θ )
= 900
Simplify
9
.
4
1002 cos(θ)sin(θ) = 900
Isolate the cosine and sine product
9
.
4
900(4 )
9
.
cos(θ )sin(θ ) =
2
100
436 Chapter 7
The left side of this equation almost looks like the result of the double angle identity for
sine: sin( θ
2 ) = 2sin(θ )cos(θ ).
By dividing both sides of the double angle identity by 2, we get
1 sin(2α) = sin(α)cos(α). Applying this to the equation above,
2
1
(
900 4 )
9
.
sin(2θ ) =
Multiply by 2
2
2
100)(
900
(
2
4 )
9
.
sin(2θ ) =
Use the inverse sine
2
100
1
−
900
(
2
)(4. )
9
2θ = sin
≈ 080
.
1
Divide by 2
1002
080
.
1
θ =
= 540
.
0
, or about 30.94 degrees
2
Power Reduction and Half Angle Identities
Another use of the cosine double angle identities is to use them in reverse to rewrite a
squared sine or cosine in terms of the double angle. Starting with one form of the cosine
double angle identity:
cos(2α) = 2cos2 (α) −1
Isolate the cosine squared term
cos(2α) +1 = 2cos2 (α)
Add 1
cos(2α) 1
cos2
+
(α) =
Divide by 2
2
cos(2α) 1
cos2
+
(α) =
This is called a power reduction identity
2
Try it Now
3. Use another form of the cosine double angle identity to prove the identity
1 cos(2α)
sin2 (α
−
) =
.
2
Example 6
Rewrite cos4 ( x) without any powers.
Since
4
cos ( x) = (cos ( x))2
2
, we can use the formula we found above
4
cos ( x) = (cos ( x))2
2
Section 7.3 Double Angle Identities 437
2
cos(2 ) +1
=
x
Square the numerator and denominator
2
(
x + )2
cos(2 ) 1
=
Expand the numerator
4
cos2 (2 x) + 2cos(2 ) +1
=
x
Split apart the fraction
4
cos2 (2 x) 2cos(2 x) 1
=
+
+
Apply the formula above to cos2 (2 x)
4
4
4
2
cos(2 2 x) 1
cos (2 x
⋅
+
) =
2
cos(4 x) +1
2
2cos(2 x) 1
=
+
+
Simplify
4
4
4
cos(4 x) 1 1
1
=
+ + cos(2 x) +
Combine the constants
8
8 2
4
cos(4 x) 1
3
=
+ cos(2 x) +
8
2
8
The cosine double angle identities can also be used in reverse for evaluating angles that
α
are half of a common angle. Building from our formula
cos(2 ) 1
cos2
+
(α) =
, if we let
2
θ
θ
θ
θ
= α
2 , then α = this identity becomes
cos( ) 1
cos2
+
=
. Taking the square
2
2
2
root, we obtain
θ
cos(θ ) +1
cos = ±
, where the sign is determined by the quadrant.
2
2
This is called a half-angle identity.
Try it Now
4. Use your results from the last Try it Now to prove the identity
θ
1− cos(θ )
sin = ±
.
2
2
Example 7
Find an exact value for cos( °
15 ).
Since 15 degrees is half of 30 degrees, we can use our result from above:
438 Chapter 7
30°
co 30
s(
)
° +1
co 15
s( )
° = cos
= ±
2
2
We can evaluate the cosine. Since 15 degrees is in the first quadrant, we need the
positive result.
3 +1
co 30
s(
)
° +1
2
=
2
2
3 1
=
+
4
2
Identities
Half-Angle Identities
θ
cos(θ ) +1
θ
−
θ
cos
1 cos( )
= ±
sin = ±
2
2
2
2
Power Reduction Identities
cos(2α) 1
α
cos2
+
(α) =
1 cos(2 )
sin2 (α
−
) =
2
2
Since these identities are easy to derive from the double-angle identities, the power
reduction and half-angle identities are not ones you should need to memorize separately.
Important Topics of This Section
Double angle identity
Power reduction identity
Half angle identity
Using identities
Simplify equations
Prove identities
Solve equations
Try it Now Answers
cos(2α ) = cos(α + α)
1. cos(α)cos(α) − sin(α)sin(α)
cos2 (α) − sin2 (α)
Section 7.3 Double Angle Identities 439
−
2.
3
co 150
s(
)
° =
2
1−cos(2α)
2
1− ( 2
2
cos (α) − sin (α))
2
2
2
−
α +
α
3. 1 cos ( ) sin ( )
2
2
2
sin (α) + sin (α)
2
2
2sin (α)
2
= sin (α)
2
1− cos(2α)
sin2 (α) =
2
1− cos(2α)
sin(α) = ±
2
θ
4. α =
2
θ
1− cos 2
θ
2
sin
= ±
2
2
θ
1− cos(θ )
sin = ±
2
2
440 Chapter 7
Section 7.3 Exercises
1. If
( x) 1
sin
= and x is in quadrant I, then find exact values for (without solving for x):
8
a. sin (2 x)
b. cos(2 x) c. tan (2 x)
2. If
( x) 2
cos
= and x is in quadrant I, then find exact values for (without solving for x):
3
a. sin (2 x)
b. cos(2 x) c. tan (2 x)
Simplify each expression.
3.
2 ( °)
2
cos 28 − sin (28 )
°
4.
2
2cos (37°) −1
5.
2
1− 2sin (17 )
°
6.
2 ( °)
2
cos 37 − sin (37 )
°
7.
2 ( x)
2
cos 9 − sin (9 x)
8.
2 ( x)
2
cos 6 − sin (6 x)
9. 4sin (8 x)cos(8 x)
10. 6sin (5 x)cos(5 x)
Solve for all solutions on the interval [0, 2π).
11. 6sin (2 t) + 9sin ( t) = 0
12. 2sin (2 t) + 3cos( t) = 0
13.
( θ )
2
9cos 2 = 9cos (θ ) − 4
14.
( α )
2
8cos 2 = 8cos (α ) −1
15. sin (2 t) = cos( t)
16. cos(2 t) = sin ( t)
17. cos(6 x) − cos(3 x) = 0
18. sin (4 x) − sin (2 x) = 0
Use a double angle, half angle, or power reduction formula to rewrite without exponents.
19.
2
cos (5 x)
20.
2
cos (6 x)
21.
4
sin (8 x)
22.
4
sin (3 x)
23.
2
4
cos x sin x
24.
4
2
cos x sin x
25. If csc( x) = 7 and 90° < x <180°, then find exact values for (without solving for x): a. sin x
x
x
b. cos
c. tan
2
2
2
26. If sec( x) = 4 and 90° < x <180°, then find exact values for (without solving for x): a. sin x
x
x
b. cos
c. tan
2
2
2
Section 7.3 Double Angle Identities 441
Prove the identity.
27. ( t −
t)2
sin
cos
= 1− sin (2 t)
28. (
x − )2
2
=
( x)
4
sin
1
cos 2 + sin x
2 tan x
29. sin (2 x)
( )
=
2
1+ tan ( x)
2sin x cos x
30. tan (2 x)
( ) ( )
=
2
2cos ( x) −1
31. cot ( x) − tan ( x) = 2cot (2 x)
sin (2θ )
32.
=
θ
1+ cos(2θ ) tan ( )
2
1− tan α
33. cos(2α )
( )
=
2
1+ tan (α )
1+ cos(2 t)
2cos( t)
34.
=
sin (2 t) − cos( t) 2sin ( t) −1
35.
( x) =
( x) 2 ( x)
3
sin 3
3sin
cos
− sin ( x)
36.
( x)
3
2
cos 3 = cos ( x) − 3sin ( x)cos( x)
442 Chapter 7
Section 7.4 Modeling Changing Amplitude and Midline
While sinusoidal functions can model a variety of behaviors, it is often necessary to
combine sinusoidal functions with linear and exponential curves to model real
applications and behaviors. We begin this section by looking at changes to the midline of
a sinusoidal function. Recall that the midline describes the middle, or average value, of
the sinusoidal function.
Changing Midlines
Example 1
A population of elk currently averages 2000 elk, and that average has been growing by
4% each year. Due to seasonal fluctuation, the population oscillates from 50 below
average in the winter up to 50 above average in the summer. Find a function that
models the number of elk after t years.
There are two components to the behavior of the elk population: the changing average,
and the oscillation. The average is an exponential growth, starting at 2000 and growing
by 4% each year. Writing a formula for this:
=
(1+ ) t = 2000(1+ 0.04) t
average initial
r
For the oscillation, since the population oscillates 50 above and below average, the
amplitude will be 50. Since it takes one year for the population to cycle, the period is 1.
π
We find the value of the horizontal stretch coefficient
original period 2
B =
=
= 2π .
new period
1
Additionally, since we weren’t told when t was first measured we will have to decide if
t = 0 corresponds to winter, or summer. If we choose winter then the shape of the
function would be a negative cosine, since it starts at the lowest value.
Putting it all together, the equation would be:
P( t) = 50
− cos(2π t) + midline
Since the midline represents the average population, we substitute in the exponential
function into the population equation to find our final equation:
( ) = 50
− cos(2π ) + 2000(1+ 0.04) t
P t
t
This is an example of changing midline – in this case an exponentially changing midline.
Changing Midline
A function of the form f ( t) = A sin( Bt) + g( t) will oscillate above and below the average given by the function g(t).
Section 7.4 Modeling Changing Amplitude and Midline 443
Changing midlines can be exponential, linear, or any other type of function. Here are
some examples:
Linear midline
Exponential midline
Quadratic midline
f ( t) = A sin ( Bt) + ( mt + b) ( ) = sin( ) + ( t
f t
A
Bt
ab )
f t = A ( Bt)
2
( )
sin
+ ( at )
Example 2
π
Find a function with linear midline of the form f t() A
= sin
t
+ mt + b that will pass
2
through the points given below.
t
0
1
2
3
f(t)
5
10
9
8