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3.5. Double Angle Identities
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3.5 Double Angle Identities
1. If sin x =
4
5
and in Quadrant II, then cosine and tangent are negative. Also, by the Pythagorean Theorem, the
p
third side is 3(b = 52 − 42 ). So, cos x = − 53 and tan x = − 34 . Using this, we can find sin 2x, cos 2x, and
tan 2x.
2 tan x
1 − tan2 x
2 · − 43
=
2
1 − − 34
cos 2x = 1 − sin2 x
2
4
= 1−2·
5
= 1−2·
sin 2x = 2 sin x cos x
3
4
= 2· ·−
5
5
24
=−
25
= 1−
=−
tan 2x =
16
25
− 83
8
7
= − ÷−
3
9
1−
8
9
= − ·−
3
7
24
=
7
=
32
25
7
25
16
9
2. This is one of the forms for cos 2x.
cos2 15◦ − sin2 15◦ = cos(15◦ · 2)
= cos 30◦
√
3
=
2
3. Step 1: Use the cosine sum formula
cos 3θ = 4 cos3 θ − 3 cos θ
cos(2θ + θ) = cos 2θ cos θ − sin 2θ sin θ
Step 2: Use double angle formulas for cos 2θ and sin 2θ
= (2 cos2 θ − 1) cos θ − (2 sin θ cos θ) sin θ
Step 3: Distribute and simplify.
= 2 cos3 θ − cos θ − 2 sin2 θ cos θ
= − cos θ(−2 cos2 θ + 2 sin2 θ + 1)
= − cos θ[−2 cos2 θ + 2(1 − cos2 θ) + 1]
→ Substitute 1 − cos2 θ for sin2 θ
= − cos θ[−2 cos2 θ + 2 − 2 cos2 θ + 1]
= − cos θ(−4 cos2 θ + 3)
= 4 cos3 θ − 3 cos θ
4. Step 1: Expand sin 2t using the double angle formula.
sin 2t − tant = tant cos 2t
2 sint cost − tant = tant cos 2t
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Chapter 3. Trigonometric Identities and Equations, Solution Key
Step 2: change tant and find a common denominator.
sint
cost
2 sint cos2 t − sint
cost
sint(2 cos2 t − 1)
cost
sint
· (2 cos2 t − 1)
cost
tant cos 2t
2 sint cost −
5.
9
If sin x = − 41
40
and in Quadrant III, then cos x = − 41
and tan x =
9
40
q
(Pythagorean Theorem, b = 412 − (−9)2 ).
So,
cos 2x = 2 cos2 x − 1
40 2
=2 −
−1
41
sin 2x = 2 sin x cos x
= 2·−
=
9
40
·−
41
41
720
1681
tan 2x =
sin 2x
cos 2x
=
3200 1681
−
1681 1681
=
720
1681
1519
1681
=
1519
1681
=
720
1519
6. Step 1: Expand sin 2x
sin 2x + sin x = 0
2 sin x cos x + sin x = 0
sin x(2 cos x + 1) = 0
Step 2: Separate and solve each for x.
2 cos x + 1 = 0
1
2
2π 4π
x= ,
3 3
cos x = −
sin x = 0
x = 0, π
or
7. Expand cos 2x and simplify
cos2 x − cos 2x = 0
cos2 x − (2 cos2 x − 1) = 0
− cos2 x + 1 = 0
cos2 x = 1
cos x = ±1
cos x = 1 when x = 0, and cos x = −1 when x = π. Therefore, the solutions are x = 0, π.
8. a. 3.429 b. 0.960 c. 0.280
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3.5. Double Angle Identities
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9. a.
2
sin 2x
2
2 csc x 2x =
2 sin x cos x
1
2 csc x 2x =
sin x cos
x
sin x
1
2 csc x 2x =
sin x
sin x cos x
sin x
2 csc x 2x = 2
sin x cos x
1
sin x
2 csc x 2x = 2 ·
sin x cos x
2 csc x 2x = csc2 x tan x
2 csc x 2x =
b.
cos4 θ − sin4 θ = (cos2 θ + sin2 θ)(cos2 θ − sin2 θ)
cos4 − sin4 θ = 1(cos2 θ − sin2 θ)
cos 2θ = cos2 θ − sin2 θ
∴ cos4 θ − sin4 θ = cos 2θ
c.
sin 2x
1 + cos 2x
sin 2x
1 + cos 2x
sin 2x
1 + cos 2x
sin 2x
1 + cos 2x
sin 2x
1 + cos 2x
sin 2x
1 + cos 2x
=
=
=
=
=
2 sin x cos x
1 + (1 − 2 sin2 x)
2 sin x cos x
2 − 2 sin2 x
2 sin x cos x
2(1 − sin2 x)
2 sin x cos x
2 cos2 x
sin x
cos x
= tan x
10. cos 2x − 1 = sin2 x
(1 − 2 sin2 x) − 1 = sin2 x
−2 sin2 x = sin2 x
0 = 3 sin2 x
0 = sin2 x
0 = sin x
x = 0, π
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Chapter 3. Trigonometric Identities and Equations, Solution Key
11.
cos 2x = cos x
2
2 cos x − 1 = cos x
2
2 cos x − cos x − 1 = 0
(2 cos x + 1)(cos x − 1) = 0
&
&
2 cos x + 1 = 0 or cos x − 1 = 0
2 cos x = −1
1
cos x = −
2
cos x = 1 when x = 0 and cos x = − 12 when x =
cos x = 1
2π
3 .
12.
2 csc 2x tan x = sec2 x
2
sin x
1
·
=
sin 2x cos x cos2 x
2
sin x
1
·
=
2 sin x cos x cos x cos2 x
1
1
=
2
cos x cos2 x
13. sin 2x − cos 2x = 1
2 sin x cos x − (1 − 2 sin2 x) = 1
2 sin x cos x − 1 + 2 sin2 x = 1
2 sin x cos x + 2 sin2 x = 2
sin x cos x + sin2 x = 1
sin x cos x = 1 − sin2 x
sin x cos x = cos2 x
p
± 1 − cos2 x cos x = cos2 x
1 − cos2 x cos2 x = cos4 x
cos2 x − cos4 x = cos4 x
cos2 x − 2 cos4 x = 0
cos2 x(1 − 2 cos2 x) = 0
.
&
1 − 2 cos2 x = 0
cos2 x = 0
cos x = 0
or
π 3π
x= ,
2 2
− 2 cos2 x = −1
1
cos2 x =
2 √
2
cos x = ±
2
π 5π
x= ,
4 4
Note: If we go back to the equation sin x cos x = cos2 x, we can see that sin x cos x must be positive or zero,
since cos2 x is always positive or zero. For this reason, sin x and cos x must have the same sign (or one of them
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3.5. Double Angle Identities
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must be zero), which means that x cannot be in the second or fourth quadrants. This is why
valid solutions.
14. Use the double angle identity for cos 2x.
sin2 x − 2 = cos 2x
sin2 x − 2 = cos 2x
sin2 x − 2 = 1 − 2 sin2 x
3 sin2 x = 3
sin2 x = 1
sin x = ±1
π 3π
x= ,
2 2
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3π
4
and
7π
4
are not