Double-Angle and Half-Angle Identities Section 5.3.

Double-Angle and Half-Angle Identities

Section 5.3

Objectives• Apply the half-angle and/or double

angle formula to simplify an expression or evaluate an angle.

• Apply a power reducing formula to simplify an expression.

Double-Angle Identities

)cos()sin(2)2sin( aaa

)(sin)(cos)2cos( 22 aaa

Half-Angle Identities

2cos1

2sin

aa

2cos1

2cos

aa

Power-Reducing Identities

2

)2cos(1sin2 a

a

2

)2cos(1cos2 a

a

Use a half-angle identity to find the exact

value of

2cos1

2sin

aa

12

sin

We will use the half-angle formula for sine

We need to find out what a is in order to use this formula.

6122

212122

aa

a

continued on next slide


value of

2

23

1

12sin

26

cos1

26sin

12

sin

We now replace a with in the formula to get6



value of

232

12sin

432

12sin

2

232

12sin

12

sin


Now all that we have left to do is determine if the answer should be positive or negative.


value of

232

12sin

12

sin


We determine which to use based on what quadrant the original angle is in. In our case, we need to know what quadrant π/12 is in. This angle fall in quadrant I. Since the sine values in quadrant I are positive, we keep the positive answer.

If

find the values of the following trigonometric functions.

23

where97

)cos( tt

)2cos( t


For this we will need the double angle formula for cosine


In order to use this formula, we will need the cos(t) and sin(t). We can use either the Pythagorean identity or right triangles to find sin(t).

If


23

where97

)cos( tt


Triangle for angle t

b

7

t

9

32

positiveislength

32

32

8149

97

2

2

222

b

b

b

b

b

If


23

where97

)cos( tt

932

)sin( t


Triangle for angle t

7

t

932

Since angle t is in quadrant III, the sine value is negative.

If


23

where97

)cos( tt

)2cos( t


Now we fill in the values for sine and cosine.

8117

)2cos(

8132

8149

)2cos(

932

97

)2cos(22

t

t

t

If


23

where97

)cos( tt

)2sin( t


For this we will need the double angle formula for cosine

)cos()sin(2)2sin( aaa We know the values of both sin(t) and cos(t) since we found them for the first part of the problem.

If


23

where97

)cos( tt

)2sin( t


813214

)2sin(

97

932

2)2sin(

t

t

Now we fill in the values for sine and cosine.

813214

)2sin( t

At this point, we can ask the question “What quadrant is the angle 2t in?”

This question can be answered by looking at the signs of the sin(2t) and cos(2t).

is positive

8117

)2cos( t is positive

and

The only quadrant where both the sine value and cosine value of an angle are positive is quadrant I.

If


23

where97

)cos( tt

2

sint


2cos1

2sin

tt


Since we know the value of cos(t), we can just plug that into the formula.

If


23

where97

)cos( tt

2

sint


297

1

2sin

2cos1

2sin

t

tt

If


23

where97

)cos( tt

2

sint


2916

2sin

297

99

2sin

t

t

If


23

where97

)cos( tt

2

sint


1816

2sin

t

Now all that we have left to do is determine if the answer should be positive or negative. In order to do this, we need to know which quadrant the angle t/2 falls in.

To do this we will need to use the information we have about the angle t.

What we need is information about t/2. If we divide each piece of the inequality, we will get t/2 in the middle of the inequality and bounds for the angle on the left and right sides.

If


23

where97

)cos( tt

23 t

2

sint


1816

2sin

t

The information that we have about the angle t is

If


23

where97

)cos( tt

2

sint


1816

2sin

t

43

22

22

3

22

23

t

t

t

Thus we see that the angle t/2 is in quadrant II.

If


23

where97

)cos( tt

2

sint


1816

2sin

t

Since the angle t/2 is in quadrant II, the sine value must be positive.

1816

2sin

t

If


23

where97

)cos( tt

2

cost


2cos1

2cos

tt


Since we know the value of cos(t), we can just plug that into the formula.

If


23

where97

)cos( tt

2

cost


297

1

2cos

2cos1

2cos

t

tt

If


23

where97

)cos( tt

2

cost


292

2cos

297

99

2cos

t

t

If


23

where97

)cos( tt

2

cost


31

91

2cos

t

Now all that we have left to do is determine if the answer should be positive or negative. In order to do this, we need to know which quadrant the angle t/2 falls in.

To do this we will need to use the information we have about the angle t.

What we need is information about t/2. If we divide each piece of the inequality, we will get t/2 in the middle of the inequality and bounds for the angle on the left and right sides.

If


23

where97

)cos( tt

23 t

2

cost


31

2cos

t

The information that we have about the angle t is

If


23

where97

)cos( tt

2

cost


31

2cos

t

43

22

22

3

22

23

t

t

t

Thus we see that the angle t/2 is in quadrant II.

If


23

where97

)cos( tt

2

cost

31

2cos

t

Since the angle t/2 is in quadrant II, the cosine value must be negative.

31

2cos

t

Use the power-reducing formula to simplify the expression

)7(sin)7(cos 44 xx

2

)2cos(1cos2 a

a

2

)2cos(1sin2 a

a


We need to use the power-reducing identity for the cosine and sine functions to do this problem.

In our problem the angle a in the formula will be 7x in our problem. We also need to rewrite our problem.

2222 )7(sin)7(cos xx


)7(sin)7(cos 44 xx


Now we just apply the identity to get:

4

)14(cos)14cos(21)14(cos)14cos(214

)14(cos)14cos(214

)14(cos)14cos(21

2)14cos(1

2)14cos(1

)7(sin)7(cos

22

22

22

2222

xxxx

xxxx

xx

xx


)7(sin)7(cos 44 xx


)14cos(4

)14cos(44

)14(cos)14cos(21)14(cos)14cos(214

)14(cos)14cos(21)14(cos)14cos(21

22

22

x

x

xxxx

xxxx

This is much simpler than the original expression and the power (exponent) is clearly reduced.


)7(sin)7(cos)7(sin)7(cos 2222 xxxx

Is there another way to simplify this without using a power-reducing formula?


The answer to this question is yes. The original expression is the difference of two squares and can be factoring into

)7(sin)7(cos 44 xx

Now you should notice that the expression in the second set of square brackets is the Pythagorean identity and thus is equal to 1.


1*)7(sin)7(cos 22 xx

)7(sin)7(cos 44 xx


Is there another way to simplify this without using a power-reducing formula?

Now you should notice that what is left is the right side of the double angle identity for cosine where the angle a is 7x.

This will allow us to rewrite the expression as

)14cos())7(2cos( xx

Double-Angle and Half-Angle Identities Section 5.3.

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