Download in the interval - Fort Thomas Independent Schools

P.o.D.
1.) Write
sin 60° cos 45° − cos 60° sin 45° as a
single trig function.
2.) Find all solutions of
2𝑠𝑖𝑛2 𝑥 − 3 sin 𝑥 = −1 on the
interval [0,2𝜋].
3.) If sin(x)=(3/5) and cos(x)=(4/5),
find tan (x).
1.) sin 15°
𝜋 𝜋 5𝜋
2.) , ,
6 2
6
3.) tan(x)=(3/4)
5.5 – Multiple Angle and Product-toSum Formulas
Learning Target: I will be able to use
multiple-angle and half-angle
formulas to evaluate trigonometric
functions.
Essential Question: How do you
rewrite trigonometric expressions
that contain functions of multiple or
half-angles, or functions that involve
square or products of trigonometric
expressions?
Double Angle Formulas:
sin 2𝑢 = 2 sin 𝑢 cos 𝑢
cos 2𝑢 = 𝑐𝑜𝑠 2 𝑢 − 𝑠𝑖𝑛2 𝑢
cos 2𝑢 = 2𝑐𝑜𝑠 2 𝑢 − 1
cos 2𝑢 = 1 − 2𝑠𝑖𝑛2 𝑢
2 tan 𝑢
tan 2𝑢 =
1 − 𝑡𝑎𝑛2 𝑢
EX: Solve cos 2𝑥 + cos 𝑥 = 0
Use a double angle identity
(2𝑐𝑜𝑠 2 𝑥 − 1) + cos 𝑥 = 0
2𝑐𝑜𝑠 2 𝑥 + cos 𝑥 − 1 = 0
Factor by De-FOIL
(2 cos 𝑥 − 1)(cos 𝑥 + 1) = 0
Set each equation equal to zero.
2 cos 𝑢 − 1 = 0
cos 𝑢 + 1 = 0
2 cos 𝑢 = 1
cos 𝑢 = −1
1
𝑢 = 𝑐𝑜𝑠 −1 (−1)
cos 𝑢 =
𝑢 = 𝜋 + 2𝑛𝜋
2
1
−1
𝑢 = 𝑐𝑜𝑠
2
𝜋 5𝜋
𝑢= ,
3 3
𝜋
𝑢 = + 2𝑛𝜋
3
5𝜋
𝑢=
+ 2𝑛𝜋
3
EX: Use a double-angle formula to
rewrite the equation
𝑔(𝑥 ) = 3 − 6𝑠𝑖𝑛2 𝑥. Then sketch the
graph of the equation over the
interval [0,2𝜋].
Begin by factoring.
𝑔(𝑥 ) = 3(1 − 2𝑠𝑖𝑛2 𝑥)
We should now recognize a doubleangle identity for cosine.
𝑔(𝑥 ) = 3 cos 2𝑥
Graph.
(show a detailed graph on the
whiteboard)
3
𝜋
5
2
EX: Use sin 𝑢 = , 0 < 𝑢 < to find
sin 2u, cos 2u, and tan 2u.
We must recognize that this is in
Quadrant I.
In order to use the double-angle
identity for sine, we need both sine
and cosine. Use a Pythagorean
Identity to find cosine.
𝑐𝑜𝑠 2 𝑥 + 𝑠𝑖𝑛2 𝑥 = 1 →
2
3
16
2
2
𝑐𝑜𝑠 𝑥 + ( ) = 1 → 𝑐𝑜𝑠 𝑥 =
5
25
4
→ cos 𝑥 =
5
Now we can apply the double-angle
identity for sine.
3 4
sin 2𝑢 = 2 sin 𝑢 cos 𝑢 = 2 ( ) ( )
5 5
24
=
25
We have three possible identities for
cos 2u. We will use 1 − 2𝑠𝑖𝑛2 𝑢 since
sine was the given value.
2
3
9
cos 2𝑢 = 1 − 2 ( ) = 1 − 2 ( )
5
25
18
7
=1−
=
25 25
We need to find tan u in order to use
the double angle identity for tangent.
sin 𝑢 3⁄5 3 5 3
tan 𝑢 =
=
= ∙ =
cos 𝑢 4⁄
5 4 4
5
Now we can use the double-angle
identity for tangent
2(3⁄4)
2 tan 𝑢
tan 2𝑢 =
=
2
1 − 𝑡𝑎𝑛 𝑢 1 − (3⁄ )2
4
3⁄
3⁄
3 16
2
2
=
=
= ∙
7⁄
9
2 7
1−
16
16
24
=
7
EX: Derive a triple-angle formula for
cos(3x).
Begin by writing 3x as a sum of 2
angles.
cos 3𝑥 = cos(2𝑥 + 𝑥)
Now apply the sum identity for
cosine.
= cos 2𝑥 cos 𝑥 − sin 2𝑥 sin 𝑥
Next, apply a double angle identity for
2x.
= (2𝑐𝑜𝑠 2 𝑥 − 1) cos 𝑥 − (2 sin 𝑥 cos 𝑥) sin 𝑥
Simplify.
= 2𝑐𝑜𝑠 3 𝑥 − cos 𝑥 − 2𝑠𝑖𝑛2 𝑥 cos 𝑥
Rewrite everything in terms of one
trig function.
= 2𝑐𝑜𝑠 3 𝑥 − cos 𝑥 − 2(1 − 𝑐𝑜𝑠 2 𝑥 ) cos 𝑥
Simplify.
= 2𝑐𝑜𝑠 3 𝑥 − cos 𝑥 − 2 cos 𝑥 + 2𝑐𝑜𝑠 3 𝑥
= 4𝑐𝑜𝑠 3 𝑥 − 3 cos 𝑥
Half-Angle Formulas:
𝑢
1 − cos 𝑢
sin = ±√
2
2
𝑢
1 + cos 𝑢
cos = ±√
2
2
𝑢 1 − cos 𝑢
sin 𝑢
tan =
=
2
sin 𝑢
1 + cos 𝑢
*Remember that the sign must always
be consistent with the quadrant.
EX: Find the exact value of cos 105°
210°
cos 105° = cos
2
1 + cos 210°
= ±√
2
−√3
2 − √3
√1 + 2
√ 2
=±
=±
2
2
√2 − √3
2 − √3
= ±√
=−
4
2
We know that our answer must be
negative since 105 degrees is in Q2.
EX: Find the exact value of tan 22.5°
tan 22.5°
45° 1 − cos 45°
= tan
=
2
sin 45°
2 − √2
√2
1−
2
2
=
=
√2
√2
2
2
2 − √2 2
4 − 2√2
=
∙
=
2
2 √2
√2
2 − √2
2 − √2 √2
=
=
∙
√2
√2
√2
2√2 − 2
=
= √2 − 1
2
2
EX: Solve 𝑐𝑜𝑠 𝑥 =
interval [0, 2𝜋]
2𝑥
𝑠𝑖𝑛
2
in the
Apply the half-angle identity for sine.
2
1 − 𝑐𝑜𝑠 𝑥
𝑐𝑜𝑠 𝑥 = (±√
)
2
2
1 − cos 𝑥
𝑐𝑜𝑠 𝑥 =
2
Simplify. (cross multiply)
2
2𝑐𝑜𝑠 2 𝑥 = 1 − cos 𝑥
Set up a quadratic.
2𝑐𝑜𝑠 2 𝑥 + cos 𝑥 − 1 = 0
Solve by factoring.
(2 cos 𝑥 − 1)(cos 𝑥 + 1) = 0
2 cos 𝑥 − 1 = 0
cos 𝑥 + 1 = 0
2 cos 𝑥 = 1
cos 𝑥 = −1
1
𝑥 = 𝑐𝑜𝑠 −1 (−1)
cos 𝑥 =
=𝜋
2
1
−1
𝑥 = 𝑐𝑜𝑠
2
𝜋 5𝜋
= ,
3 3
*Remember, we can always confirm
our answers graphically.
Product to Sum Formulas:
1
sin 𝑢 sin 𝑣 = [cos(𝑢 − 𝑣) − cos(𝑢 + 𝑣)]
2
1
cos 𝑢 cos 𝑣 = [cos(𝑢 − 𝑣) + cos(𝑢 + 𝑣)]
2
1
sin 𝑢 cos 𝑣 = [sin(𝑢 + 𝑣) + sin(𝑢 − 𝑣)]
2
1
cos 𝑢 sin 𝑣 = [sin(𝑢 + 𝑣) − sin(𝑢 − 𝑣)]
2
EX: Rewrite sin 5𝜃 cos 3𝜃 as a sum or
difference.
1
1
(
)
= [sin 5𝜃 + 3𝜃 + sin(5𝜃 − 3𝜃)] = [sin 8𝜃 + sin 2𝜃]
2
2
Sum to Product Formulas:
𝑢+𝑣
𝑢−𝑣
sin 𝑢 + sin 𝑣 = 2 sin (
) cos (
)
2
2
𝑢+𝑣
𝑢−𝑣
sin 𝑢 − sin 𝑣 = 2 cos (
) sin (
)
2
2
𝑢+𝑣
𝑢−𝑣
cos 𝑢 + cos 𝑣 = 2 cos (
) cos (
)
2
2
𝑢+𝑣
𝑢−𝑣
cos 𝑢 − cos 𝑣 = −2 sin (
) sin (
)
2
2
EX: Find the exact value of
sin 195° + sin 105°
195° + 105°
195° − 105°
= 2 sin (
) cos (
)
2
2
300°
90°
= 2 sin
cos
2
2
1 √2
= 2 sin 150° cos 45° = 2 ( ) ( )
2
2
√2
=
2
EX: Solve sin(4x)-sin(2x)=0 on the
interval [0,2𝜋]
*Let’s solve this graphically.
𝜋 𝜋 5𝜋
7𝜋 9𝜋 11𝜋
𝑥 = 0, , ,
, 𝜋,
,
,
6 2 6
6 6 6
EX: Verify the identity
sin 6𝑥+sin 4𝑥
cos 6𝑥+cos 4𝑥
= tan 5𝑥
6𝑥 + 4𝑥
6𝑥 − 4𝑥
2 sin (
) cos (
)
2
2
=
6𝑥 + 4𝑥
6𝑥 − 4𝑥
2 cos (
) cos (
)
2
2
2 sin 5𝑥 cos 𝑥
=
= tan 5𝑥
2 cos 5𝑥 cos 𝑥
EX: Ignoring air resistance, the range of a
projectile fired at an angle 𝜃 with the
horizontal and with an initial velocity of
𝑣0 feet per second is given by 𝑟 =
1
16
𝑣0 2 sin 𝜃 cos 𝜃 where r is the
horizontal distance (in feet) that the
projectile will travel. A place kicker for a
football team can kick a football from
ground level with an initial velocity of 78
feet per second.
a.) At what angle must the player kick
the football so that it travels 188
feet?
1
(78)2 (sin 𝜃 cos 𝜃)
188 =
16
1
(78)2 (2 sin 𝜃 cos 𝜃)
188 =
32
6084
188 =
(sin 2𝜃)
32
1504
= sin 2𝜃
1521
1504
−1
𝑠𝑖𝑛
= 2𝜃
1521
81.43°, 98.57° = 2𝜃
40.715°, 49.285° = 𝜃
b.) For what angle is the horizontal
distance the football travels a
maximum?
Let’s solve this graphically.
The ball will travel a maximum distance
when it reaches its maximum range.
Therefore, the best angle is 45 degrees.
Upon completion of this lesson, you
should be able to:
1. Simplify using Double angle and
Half angle Identities.
For more information, visit
http://www.sosmath.com/trig/douangl/douangl.html
HW
Pg.415
6-90 6ths, 99, 119.
Quiz 5.4-5.5 tomorrow