3.5 DoubleAngleIdentities - All-in-One High School

3.5. Double Angle Identities



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

= ? ¡Â?

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?

¡Ì

3

=

2

3. Step 1: Use the cosine sum formula

cos 3¦È = 4 cos3 ¦È ? 3 cos ¦È

cos(2¦È + ¦È) = cos 2¦È cos ¦È ? sin 2¦È sin ¦È

Step 2: Use double angle formulas for cos 2¦È and sin 2¦È

= (2 cos2 ¦È ? 1) cos ¦È ? (2 sin ¦È cos ¦È) sin ¦È

Step 3: Distribute and simplify.

= 2 cos3 ¦È ? cos ¦È ? 2 sin2 ¦È cos ¦È

= ? cos ¦È(?2 cos2 ¦È + 2 sin2 ¦È + 1)

= ? cos ¦È[?2 cos2 ¦È + 2(1 ? cos2 ¦È) + 1]

¡ú Substitute 1 ? cos2 ¦È for sin2 ¦È

= ? cos ¦È[?2 cos2 ¦È + 2 ? 2 cos2 ¦È + 1]

= ? cos ¦È(?4 cos2 ¦È + 3)

= 4 cos3 ¦È ? 3 cos ¦È

4. Step 1: Expand sin 2t using the double angle formula.

sin 2t ? tant = tant cos 2t

2 sint cost ? tant = tant cos 2t

50



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

2¦Ð 4¦Ð

x= ,

3 3

cos x = ?

sin x = 0

x = 0, ¦Ð

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 = ¡À1

cos x = 1 when x = 0, and cos x = ?1 when x = ¦Ð. Therefore, the solutions are x = 0, ¦Ð.

8. a. 3.429 b. 0.960 c. 0.280

51

3.5. Double Angle Identities



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 ¦È ? sin4 ¦È = (cos2 ¦È + sin2 ¦È)(cos2 ¦È ? sin2 ¦È)

cos4 ? sin4 ¦È = 1(cos2 ¦È ? sin2 ¦È)

cos 2¦È = cos2 ¦È ? sin2 ¦È

¡à cos4 ¦È ? sin4 ¦È = cos 2¦È

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, ¦Ð

52



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

2¦Ð

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



¡À 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

¦Ð 3¦Ð

x= ,

2 2

? 2 cos2 x = ?1

1

cos2 x =

2 ¡Ì

2

cos x = ¡À

2

¦Ð 5¦Ð

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

53

3.5. Double Angle Identities



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 = ¡À1

¦Ð 3¦Ð

x= ,

2 2

54

3¦Ð

4

and

7¦Ð

4

are not

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