3.5 DoubleAngleIdentities - All-in-One High School

[Pages:5]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

third side is 3(b =

52 - 42).

So,

cos x

=

-

3 5

and

tan x

=

-

4 3

.

Using this, we can find sin 2x, cos 2x, and

tan 2x.

sin 2x = 2 sin x cos x

= 2? 4 ?-3 55 24

=- 25

cos 2x = 1 - sin2 x

42 = 1-2?

5

= 1 - 2 ? 16 25

= 1 - 32 25

7 =-

25

2 tan x tan 2x = 1 - tan2 x

=

2

?

-

4 3

1-

-

4 3

2

=

-

8 3

1

-

16 9

= -8 3

?-7 9

= -8 ?-9 37

24 =

7

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 3 = 4 cos3 - 3 cos cos(2 + ) = cos 2 cos - sin 2 sin

Step 2: Use double angle formulas for cos 2 and sin 2 = (2 cos2 - 1) cos - (2 sin cos ) sin

Step 3: Distribute and simplify.

= 2 cos3 - cos - 2 sin2 cos = - cos (-2 cos2 + 2 sin2 + 1) = - cos [-2 cos2 + 2(1 - cos2 ) + 1] = - cos [-2 cos2 + 2 - 2 cos2 + 1] = - cos (-4 cos2 + 3) = 4 cos3 - 3 cos

Substitute 1 - cos2 for sin2

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.

sin t 2 sint cost -

cos t 2 sint cos2 t - sint

cos t sint(2 cos2 t - 1)

cos t sint ? (2 cos2 t - 1) cos t

tant cos 2t

5.

If

sin

x

=

-

9 41

and

in

Quadrant

III,

then

cos

x

=

-

40 41

and

tan

x

=

9 40

(Pythagorean Theorem, b =

412 - (-9)2).

So,

sin 2x = 2 sin x cos x 9 40

=2?- ?- 41 41

720 =

1681 6. Step 1: Expand sin 2x

cos 2x = 2 cos2 x - 1 40 2

=2 - -1 41

3200 1681 =-

1681 1681 1519 = 1681

sin 2x tan 2x =

cos 2x

720

=

1681 1519

1681

720 =

1519

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.

sin x = 0

x = 0,

or

7. Expand cos 2x and simplify

2 cos x + 1 = 0

1 cos x = -

2 x = 2 , 4

33

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 = . Therefore, the solutions are x = 0, . 8. a. 3.429 b. 0.960 c. 0.280

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3.5. Double Angle Identities

9. a.

2 2 csc x 2x =

sin 2x 2

2 csc x 2x = 2 sin x cos x 1

2 csc x 2x = sin x cos x

2 csc x 2x = sin x

1

sin x sin x cos x

sin x 2 csc x 2x = sin2 x cos x

1 sin x 2 csc x 2x = sin2 x ? cos x 2 csc x 2x = csc2 x tan x

b.

cos4 - sin4 = (cos2 + sin2 )(cos2 - sin2 ) cos4 - sin4 = 1(cos2 - sin2 ) cos 2 = cos2 - sin2

cos4 - sin4 = cos 2

c. 10. cos 2x - 1 = sin2 x

sin 2x

2 sin x cos x

1 + cos 2x = 1 + (1 - 2 sin2 x)

sin 2x 2 sin x cos x 1 + cos 2x = 2 - 2 sin2 x

sin 2x 2 sin x cos x 1 + cos 2x = 2(1 - sin2 x)

sin 2x 2 sin x cos x

= 1 + cos 2x

2 cos2 x

sin 2x sin x =

1 + cos 2x cos x

sin 2x = tan x

1 + cos 2x

(1 - 2 sin2 x) - 1 = sin2 x -2 sin2 x = sin2 x 0 = 3 sin2 x 0 = sin2 x 0 = sin x x = 0,

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Chapter 3. Trigonometric Identities and Equations, Solution Key

11.

cos 2x = cos x 2 cos2 x - 1 = cos x 2 cos2 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

cos x = 1

cos x = - 1 2

cos x

=

1

when

x

=

0

and

cos x

=

-

1 2

when

x

=

2 3

.

12.

13. sin 2x - cos 2x = 1

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

cos2 x = cos2 x

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

? 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

- 2 cos2 x = -1

cos x = 0 or x = , 3 22

cos2 x = 1 2 2

cos x = ? 2

x = , 5 44

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



must

be

zero),

which

means

that

x

cannot

be

in

the

second

or

fourth

quadrants.

This

is

why

3 4

and

7 4

are

not

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 x = , 3 22

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