10.3 Double-Angle and Half-Angle Formulas Objective 1)To derive and apply double-angle and half-angle formula for sine, cosine, and tangent. 2)To use the

10.3 Double-Angle and Half-Angle Formulas

Objective 1) To derive and apply double-angle and

half-angle formula for sine, cosine, and tangent.

2) To use the double-angle and half-angle formulas to simplify trigonometric expressions and verify the trigonometric identities.

3) To apply the double-angle and half-angle formulas in solving some trigonometric equations.

Question Does sin2 = 2sin ?

Use some specific value to verify the above equality is correct or wrong.

If take = . The above equality is correct.

If take = /2, then

sin2 = sin = 0

and

2sin = 2 sin(/2) = 2.

Therefore, sin2 2sin . What is sin2 ?

Example. An oscilloscope often displays a sawtooth curve. This curve can be approximated by sinusoidal curves of varying periods and amplitudes. A first approximation to the sawtooth curve is given by

-4 -2 2 4

-4

-2

2

4

-4 -2 2 4

-4

-2

2

4

It has the exact graph as

y = 2sin(x)cos3(x)

This example tell us that we need to explore and derive the double-angle formula

The double-angle formulas can be obtained very easily from the Sum and Difference formulas for sine and cosine. Recall

sin( ) sin cos cos sin


cos( ) cos cos sin sin


(1)

(2)

(3)

(4)

To derive the double-angle formula for sine, we just need set = in formula (1)





(1)

(2)

(3)

(4)

We get

sin2 = sin cos + cos sin = 2sin cos

sin2 = 2sin cos

To derive the double-angle formula for cosine, we just need set = in formula (3)





(1)

(2)

(3)

(4)

We getcos2 = cos2 – sin2 = (1 – sin2 ) – sin2

= 1 – 2sin2 = cos2 – ( 1 – cos2 ) = 2 cos2 – 1

cos2 = cos2 – sin2 = 1 – 2sin2 = 2cos2 – 1

2

tan tan 2 tantan 2

1 tan tan 1 tan

To derive the double-angle formula for tangent, we just need set = in tangent of sum-angle formula.

tan tantan( )

1 tan tan

2

2 tantan 2

1 tan

Summary of Double-Angle Formulas

sin2 = 2sin cos

cos2 = cos2 – sin2 = 1 – 2sin2

= 2cos2 – 1

2

2 tantan 2

1 tan

You can see from these four formulas that the double-angle formula can be interpreted as a function of same (or co)trig function of single-angle but with higher power. It seems like

F(double-angle) = G(single-angle) higher power.

Double-angle formula for cosine can be expressed in more than one way.

Example 1: If , find sin2x given that

sin x < 0.3

1cos x

2 2 1sin cos 1, cos ,

3x x x

22 1

sin 1 3

x

2 8sin

9x

9

24

3

1

3

222cossin22sin

xxx

[Solution] Recall the Pythagorean relationship

2 2Since sin 0, sin

3x x

Example 2: If , and 0 < < /2, find

sin2 , cos2, and tan2 .

4sin

5

2 2 4sin cos 1, sin , and 0 <

5 2

3 cos ,

5

4 3 24sin 2 2sin cos 2

5 5 25

sin 2 24 / 25 24tan 2

cos 2 7 / 25 7

[Solution] Recall the Pythagorean relationship

22 4 7

Since cos 2 1 2sin , cos 2 1 25 25

This is the book example on P. 381. My way is slightly different.

Example 3: Find the exact value of

(a) 2sin67o30’cos67o30’ (b) 2 25sin sin

12 12

sin 2 2sin cos

o o o o 2 2sin 67 30 'cos67 30 ' sin(2 67 30 ') sin135

2

5(b) Since sin cos cos ,

12 2 12 12

2 25 5 5 5 3cos sin cos 2 cos

12 12 12 6 2

[Solution] (a) Recall the double-angle formula for sine:

2 2 2 2

Then

5 5 5 sin sin sin cos

12 12 12 12

Example 4: Prove the identity

sin cos4 4

2sec 2sin cos

4 4

x xx

x x

sin cos sin cos cos sin4 4 4 4 4 4

sin cos sin cos4 4 4 4

x x x x x x

x x x x

sin4 4

sin cos4 4

x x

x x

[Proof] Start from the left side

2sin2

2sin cos4 4

x x

2

sin 22

x

2

cos 2x 2sec 2x

Practice: Simplify the following:

2 2 2

2 2 2

4 4

(a) cos sin (b) sec cos 2cos2 2

3(c) sin 2cos + cos (d) 2sin sin( )

2 2 2

(e) cos sin

x x x x x

x x

(a) cos(2 ) (b) cos 2

(c) cos (d) sin 2

(e) cos 2

x x

Key

Half-Angle FormulasAfter we get the double-angle formula for sine, cosine and tangent, if we make backwards substitution in cosine double-angle formulas, we can get half-angle formulas easily.

cos2 = 1 – 2sin2 = 2cos2 – 1

We let = 2, then = /2, so the above formulas are:

cos = 1 – 2sin2 /2 (1)

cos = 2cos2 /2 – 1 (2)

The above two quadratic equations are with respect to sine and cosine of /2. Solve these two equations, we get

Half-Angle FormulasAfter we get the double-angle formula for sine, cosine and tangent, if we make backwards substitution in cosine double-angle formulas, we can get half-angle formulas easily.

cos2 = 1 – 2sin2 = 2cos2 – 1

We let = 2, then = /2, so the above formulas are:

cos = 1 – 2sin2 /2 (1)

cos = 2cos2 /2 – 1 (2)

The above two quadratic equations are with respect to sine and cosine of /2. Solve these two equations, we get

22sin 1 cos 1 cos

sin2 2

2

Therefore, the tan /2 can be directly derived from half-angle of sine and cosine above:

1 cossin

221 coscos

2

1 cost

2 1 cos

2

an

In textbook, the statement on top of P. 383 addresses the alternative half-angle formulas for tangent can be derived by simplifying the radical expression of tan /2. HOW?

22cos 1 cos 1 cos

cos2 2

2

In my personal opinion, that statement is not workable. Actually, the tan /2 alternative formulas without the ambiguous sign can be derived as following:

sin 2sin cos2 2tan

22

cos 2cos cos2 2 2

Notice the Pythagorean relationship: sin2 = 1 – cos2,

2

sin

2cos2

sin

1 cos

or, sin2 = (1 – cos )(1 + cos )

Dividing sin (1 + cos ) at both sides, we obtain:sin

1 co

os

s

1 c

sin

sin 1 costan

2 1 cos sin

Or, the other alternative formula for tan /2 without the ambiguous sign can be derived as following:

sin 2sin sin2 2tan

22

cos 2cos sin2 2 2

You can see that in the trigonometry, there is more than one way to get same or an equivalent expression.

22sin2

sin

1 cos

sin

One of the alternative half-angle formulas for tan /2 can be derived in a very nice geometric way (pretended that is an acute angle):

sintan

2 1 cos

BC

AC

In right triangles BOC, OC = cos, BC = sin.

Then

y

x0 /2

1 sin

cos1A

B

CIn right triangles ABC, AC = 1 + cos.

Again!!! You can see that in the trigonometry, there is more than one way to get the same or an equivalent expression.

Summary of Half-Angle Formulas

1 cossin

2 2

1 coscos

2 2

sintan

2 1 cos

1 cos

sin

1 costan

2 1 cos

(1)

(2)

(3)

(4)

(5)

Example 5: Find the exact value of 5

cos8

5cos 0

8

51 cos5 4cos

8 2

21

2

2

[Solution] Since 5/8 is in the 2nd Quadrant,

2 2 2 2

4 2


sin15o, cos15o, and tan15o

oo

311 cos30 2 3 2 32sin15 = = =

2 2 4 2

22( 3 1) ( 3 1) 2

16 4

6 2

4

[Solution] Apply the half-angle formula:

22(2 3) 4 2 3 3 2 3 1 ( 3 1)

8 8 8 8



oo

311 cos30 2 3 2 32cos15 = = =

2 2 4 2

22( 3 1) ( 3 1) 2

16 4

6 2

4


22(2 3) 4 2 3 3 2 3 1 ( 3 1)

8 8 8 8



oo

o

1sin 30 2tan15 = =

1 cos30 31

2


1 2 32 3

2 3 2 3

1 =

2 3

Practice: Find the exact values for:o o o(a) sin22.5 (b) cos22.5 (c) tan 22.5

2 2 2 2(a) (b) (c) 2 1

2 2

Key

Example 7: Write 1 + cos + sin as product.

21 cos 2cos ,2

[Solution] Notice the double-angle and half-angle formula:

Therefore,

sin 2sin cos2 2

21 cos sin 2cos 2sin cos2 2 2

2cos cos sin2 2 2

Example 8: If is in the 2nd quadrant, and

tan = – 4/3. Find

(a) (b) (c)

3cos = ,

5

[Solution] Since is in the 2nd quadrant, tan = – 4/3, so

4 2 2

k k

2 22

k k

sin2

cos

2

tan

2

r2 = 42 + (– 3)2 = 25, r = 5.

Therefore,

/2

y

x0–3

4 5 /2

4sin = .

5

If k is odd, then /2 is in 3rd quadrant. If k is even, then /2 is in 1st quadrant.


tan = – 4/3. Find

(a) (b) (c)

1 cos 1 ( 3 / 5)sin =

2 2 2

4 =

5

sin2

cos

2

tan

2

(a)

/2

y

x0–3

4 5 /2

[Solution] If k is odd, then /2 is in 3rd quadrant. So

2 5 =

5

1 cos 1 ( 3 / 5)cos =

2 2 2

1 =

5

(b)

5 =

5


tan = – 4/3. Find

(a) (b) (c)

sin 4 / 5tan =

2 1 cos 1 ( 3 / 5)

=2

sin2

cos

2

tan

2

(c)

/2

y

x0–3

4 5 /2

[Solution] If k is odd, then /2 is in 3rd quadrant. So

If k is even, then /2 is in 1st quadrant.

1 cos 1 ( 3 / 5)sin =

2 2 2

4 =

5

(a)

2 5 =

5


tan = – 4/3. Find

(a) (b) (c)sin2

cos

2

tan

2

/2

y

x0–3

4 5 /2

[Solution] If k is even, then /2 is in 1st quadrant.

1 cos 1 ( 3 / 5)cos =

2 2 2

1 =

5

(b)

5 =

5

sin 4 / 5tan =

2 1 cos 1 ( 3 / 5)

=2(c)

PracticeSimplify the expression by using a double-angle or half-

angle formula.

1.

2.

3.

4. Find if and u is in

quadrant IV.

2sin 35 sin 35

2 2cos sin2 2

1 cos 4

2

sin , cos , and tan ,2 2 2

u u u 2cos

3u

o(1) 1 cos70 (2) cos (3) sin 2

KeyPractice

3(4) 2 2 2 ,

2k u k k

Z

3,

4 2

uk k k

Z

is in 2nd Quad., if is even.2

uk

is in 4th Quad., if is odd.2

uk

If is even

6 30 5sin cos tan

2 6 2 6 2 5

k

u u u

KeyPractice

If is odd

6 30 5sin cos tan

2 6 2 6 2 5

k

u u u

Power-Reducing FormulaPower-Reducing Formula

From the double-angle formulas, we have learned that the expression of double-angle is ended in angle-halving and power-raising.

sin2 = 2sin cos

cos2 = cos2 – sin2 = 1 – 2sin2

= 2cos2 – 1

2

2 tantan 2

1 tan

Angle-halving2nd power

Power-Reducing FormulaPower-Reducing Formula

2

2cos1sin 2 x

x

2

2cos1cos2 x

x

x

xx

2cos1

2cos1tan2

From the half-angle formulas, we have learned that the expression of half-angle is ended in angle-doubling and power-reducing.

Angle-doubling 1st power

22 cossin3

Example 9: Reduce the power of:

[Solution 1] The expression has power of 4. Apply the power-reducing formula, we have

2 2 1 cos 2 1 cos 23sin cos 3

2 2

3(1 cos 2 ) (1 cos 2 )

4

2 23 3(1 cos 2 ) sin 2

4 4

3 1 cos 4 3(1 cos 4 )

4 2 8

22 cossin3

Example 9: Reduce the power of:

[Solution 2] The expression has power of 4. Apply the double-angle formula reversely, we have

2 2 2 233sin cos 4sin cos

4

23(2sin cos )

4

2 23 3(sin 2 ) sin 2

4 4

3 1 cos 4 3(1 cos 4 )

4 2 8

02sincos2 xxExample 10: Solve the trig equation:

Solving More Difficult Trigonometric EquationSolving More Difficult Trigonometric Equation

[Solution] Like solving any polynomial equation, it will be much easier to solve the trigonometric equation in a factor form. We start to factor the equation.

2cos sin 2 2cos 2sin cos 2cos (1 sin )x x x x x x x

Therefore, the original equation becomes:

2cos (1 sin ) 0.x x

After we learned the double-angle and half-angle and power-reducing formulas, we can solve some more difficult trigonometric equation.

So,cos 0, or 1 sin 0.x x

02sincos2 xxExample 10: Solve the trig equation:

Solving More Difficult Trigonometric EquationSolving More Difficult Trigonometric Equation

[Solution]

cos 0, , .2

x x k k Z

1 sin 0, sin 1 2 , 2

x x x k k Z

cos 0, or 1 sin 0.x x

So the solutions are

, or 2 , 2 2

x k x k k Z

2cos2sin2 22 x

x

Example 11: Solve the trig equation:

[Solution] Rewrite the expression at the left,

22cos 1 cos2

xx

2 2 22 sin 1 1 sin 1 cosx x x

Power-reducing the expression at the right,

So we end up with an trig equation:21 cos 1 cosx x cos (cos 1) 0x x

cos 0, or cos 1.x x

, or 2 (2 1) , 2

x k x k k k Z

sin 2 sin 6 0x x Example 12: Solve the trig equation:

[Solution] Rewrite sum to product

sin 4 cos( 2 ) 0x x

2 6 2 62sin cos 0

2 2

x x x x

Then

So,sin 4 0, or cos 2 0.x x

4 , or 2 , 2

x k x k k Z

sin 4 cos 2 0x x

, or , 4 2 4

k kx x k

Z

Linear Combination of Sine and CosineLinear Combination of Sine and Cosine

A linear combination of sine and cosine is always equivalent to a sine or cosine plus a non-zero phase. In the next chapter we will use this skill to switch an algebraic expression of a complex number to the polar form. In the Calculus class you will find this skill is very useful while you deal with an integral.

What is a linear combination of sine and cosine?

A linear combination of sine and cosine is of the form

a sin + b cos , where a, b R

How can we change this linear combination to a sine or cosine?

Here is what we will do:


Step 1 Find r: 2 2 .r a b

Step 2 Factor out r in the linear combination

sin cosa b 2 2

2 2 2 2sin cos

a ba b

a b a b

Step 3 Use the segment joining point (a, b) and the origin as terminal side, the angle formed with positive x-axis is the angle we are looking for as a phase, denoted as ϕ. Therefore,

2 2 2 2cos , and sin

a b

a b a b


Step 4 Rewrite the linear combination as

2 2

2 2 2 2sin csin c s so o

a ba b

aa b

b a b

where ϕ is the angle formed by segment joining point (a, b) and the origin and positive x-axis.

2 2 cos sin sin cosa b

2 2 sin cos cos sina b

2 2 sin( )a b

2 2sin cos sin( )a b a b

2 sin cos 2x x Example 13: Solve the trig equation:

[Solution] Rewrite the linear combination

2 13 sin cos 2

3 3x x

2( 2) 1 3r

Then factor out r

Let2 1

cos , and sin .3 3

3 sin cos cos sin 2.x x

3 sin( ) 2x 2 6

sin( )33

x

2 sin cos 2x x Example 13: Solve the trig equation:

[Solution]

Then the general solution for sin(x – ϕ) above is

3 sin( ) 2x 2 6

sin( )33

x

6( 1) arcsin ,

3kx k k

Z

Since ϕ can be expressed as3

arcsin ,3

6 3( 1) arcsin arcsin ,

3 3kx k k

Z

Therefore, the general solution is

Assignment

P. 383 #1 – 18, 19 – 25 (odd), 31, 33, 37

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10.3 Double-Angle and Half-Angle Formulas Objective 1)To derive and apply double-angle and half-angle formula for sine, cosine, and tangent. 2)To use the