Half-Angle Identities

Half-Angle Identities

In order to find half-angle equations you must be able to use many trigonometric equations and be able to use a unit circle to find values of cosine, sine, and tangent (see page #1). Here are some equations you will need to know:

Compound Angles: 1. Sin(A + B) = sinAcosB + cosAsinB

2. Sin(A - B) = sinAcosB - cosAsinB

3. cos(A + B) = cosAcosB – sinAsinB

4. cos(A - B) = cosAcosB + sinAsinB

5. tan(A +B) = tanA + tanB

1 – tanAtanB

6. tan(A - B) = tanA - tanB

1 + tanAtanB

Double Angles: 1. Sin(2A) = 2sinAcosA

2. cos(2A) = Cos2A - sin2A

3. cos(2A) = 2Cos2A – 1

4. cos(2A) = 1 - 2sin2A

5. tan(2A) = 2tanA

1 – tan2A

Trig Identities: 1. sin2A + Cos2A = 1

2. tan2A + 1 = Sec2A

3. Cot2A +1 = Csc2A

4. Tan = Sin 5. Cot = 1 6. Sec = 1 7. Csc = 1

Cos Tan Cos Sin

By using the double angle equations 3 and 4 we may derive the equations for the half-angles.

Start with double angle Identity 3:

Cos(2A) = 2Cos2A + 1

Rearrange the equation to isolate cos2A by moving the -1 to the left and dividing both sides by 2:

Cos2A = 1 + Cos(2A)

Since we are finding the HALF-angles we must find HALF of A. To do this you must replace every A with A:

Cos2A = 1 + Cos(2A/2)

2 2

Therefore Half-angle Identity 1 is:

Cos2A = 1 + CosA

2 2

To find the half-angle identity for sine follow the same procedure as in the proof above only begin with Double angle Identity 4:

Cos(2A) = 1 - 2sin2A

Rearrange: Sin2A = 1 - Cos(2A)

Replace A with A: Sin2(A) = 1 - Cos(2A/2)

2 2 2

Therefore Half-angle Identity 2 is:

Sin2A = 1 – CosA

2 2

Now that we have the half-angle identities for cosine and sine we can find the two half-angle identities for tangent. Using Trig Identity 4 we see that Tan2x = Sin2x so we can use this to find Tan A/2. Cos2x

Tan2(A/2) = Sin2(A/2) which equals Tan2(A/2) = 1 - CosA

Cos2(A/2) 1 + CosA

These are Half-angle Identities 3 and 4.

By multiplying both the numerator and the denominator of the right side of the equation by (1 - CosA) and taking the square root of both sides of the equation and then doing the same thing only using (1 + CosA) you arrive at two more Identities for the half-angles of tangent:

Half-angle identity 5: TanA/2 = 1 + CosA Half-angle identity 6: TanA/2 = SinA

SinA 1 + CosA

Now that the half-angle identities have been derived we can begin to find the product to sum identities. There are four product to sum formulas.

For the first add Sin(A + B) and Sin(A - B) by using the compound angle identities l and 2:

= Sin(A + B) + sin(A - B)

= [(sinAcosB + cosAsinB) + (sinAcosB - cosAsinB)]

= 2sinAcosB

Therefore product to sum identity 1 is: 2SinACosB = Sin(A + B) + Sin(A - B)

The next equation is:

Sin(A+B) - sin(A-B)

= [(sinAcosB + cosAsinB) - (sinAcosB - cosAsinB)]

= sinAcosB + cosAsinB - sinAcosB + cosAsinB

= 2cosAsinB

Therefore product to sum identity 2 is: 2cosAsinB = sin(A + B) - sin(A - B)

The remaining two equations are derived in the same way:

Cos(A+B) + Cos(A-B)

= [(cosAcosB - sinAsinB) + (cosAcosB + sinAsinB)

= 2cosAcosB

Therefore product to sum identity 3 is: 2cosAcosB = cos(A + B) + cos(A - B)

cos(A+B) - cos(A-B)

= [(cosAcosB - sinAsinB) - (cosAcosB + sinAsinB)] = -2sinAsinB

Therefore product to sum identity 4 is: -2sinAsinB = cos(A + B) - cos(A - B)

Now that we have solved the product-to-sum equations we can find the Sum to product equations.

To solve for these four other equations we must note a change that will help us solve the formula:

let (A+B) = C and let (A-B) = D

Using these and the equations above: (A+B) + (A-B) = C+D

Which equals: A = C+D

And: (A+B) - (A-B) = C-D

Which equals: B = C-D

Now using the product to sum formulas and substitute C+D for A and C-D for B on the right-hand side of the equation, and the values for A and B on the left-hand side of the equation.

Example: 2sinAcosB = sin(A+B) + sin(A-B)

would become

2SinC+DCosC-D = SinC + SinD

2 2

Using that example above and applying the same steps to the other product to sum equations we can come up with the four sum to product equations:

1. SinC + SinD = 2 SinC+DCosC-D

2 2

2. SinC - SinD = 2CosC+DSinC-D

2 2

3. CosC + CosD = 2CosC+DCosC-D

2 2

4. CosC - CosD = -2SinC+DSinC-D

2 2

Now we have found all of the equations needed to solve for half-angles. The following are some examples of problems you will need to use these twelve new equations in order to solve the problem.

Example 1: Find the exact value of Cos(15°)

Since 15° = 30°/2 use the half-angle formula for Cos(A/2) with A= 30°.

Also, because 15° is in the quadrant I Cos(15°) is positive

(always keep your unit circle positions in mind when answering questions).

Cos(15°) = Cos(30°/2) = 1 + Cos30°

(the same as taking the square root on both sides of the Cos230°/2)

= 1 + √(3 /2)

Example 2: Prove that SinX Tan(X/2) + 2cosx = 2cos2(X/2)

In questions such as this, pick one side to work with and leave the other side the way it is. In this example we are going to work with the left side.

= sinx tan(x/2) + 2cosx

= sinx (1-cosx) +2cosx substitute for tan half-angle

sinx

= 1-cosx +2cosx expand and simplify

= 1+cosx rearrange cosine half angle Identity

= 2Cos2(x/2) left side = right side

Example 3: Express 2sin3xcos5x as a sum using only sines and cosines in your answer.

In the equation given A would be 3x and B would be 5x. Use the first product to sum identity.

2sin3xcos5x = sin(3x + 5x) + sin (3x - 5x)

= sin (8x) + sin (-2x)

= sin 8x - sin 2x

(Since sin(-x) = -sin(x) your answer is sin 8x - sin 2x.)

Example 4: Express cos3x + cos2x as a product of sines and cosines.

If cos3x + cos2x, then C = 3x and D = 2x. Use the third sum to product identity.

Cos3x + cos 2x = 2cos3x + 2x cos3x - 2x = 2cos 5x cos x

2 2 2 2

Cast Rule:

Only Sine is positive All (Sine, Cosine, Tangent, etc) are

in this quadrant. positive in this quadrant.

IV S A І

III T C II

Only Tangent is positive Only Cosine is positive in this

in this quadrant. quadrant.

Unit Circle:

Now, here are some problems to try!

Practice Problems

1. Find the value of each trig function using half-angle identities:

a) sin 22.5° b) tan 7π c) cos 165° d) sin (-3π)

8 8

2. Prove that the left side equals the right side.

a) cos4x - sin4x = cos 2x

b) cos22x - sin22x = cos 4x

c) cosx = 1 - tan2(x/2)

1 + tan2(x/2)

d) sin3x - cos3 = 2

sinx cosx

e) tan3x = 3tanx – tan3x

1 – 3tan2x

3. Express each product as a sum using only sines and cosines in your answer.

a) 2sin4xsin2x

b) 2cos3xcos5x

c) 2sinxsin2x

d)2sin 3x cosx

2 2

4. Express each sum as a product.

a) sin4x - sin2x b) cos2x + cos 4x c) sinx + sin3x

d) cosx - cos3x e) sin6x + sin2x f) cos5x - cos 3x

2 2

Test Half Angle Identities

Test

Questions:

Part I

Use a brainy choice of A and B, ( such as 30º, 45º, 60º, 90º, etc.) and any of the properties you need to find exact values of the following. Express the answers in simple radical form.

1. 2.

Sin 165º Tan (-15º)

Part II

Prove the following equations that they are identities.

3. 4.

1- Cos (2A)___ = Tan (A/2) 2 Sin²(A/2) + CosA = 1

2 Sin A + Sin (2A)

5. 6.

Tan ½A + Cot ½A = 2 CscA SinB Tan (B/2) + 2 CosB = 2 Cos²(B/2)

CosB – Tan(B/2) Sin(2B) – Cos(2B) = 2Sin²(B/2)

Tan (B/2) = CscB – CotB

Part III

Transform the indicated Sum (or Difference) into a product of sins and cosines of positive arguments.

9. 10.

Cos80º Cos20º + Sin80ºSin20º Cos 2.4x – Cos 4.4x

11. 12.

2 Cos (3.8x) Sin (4.1x) Sin 30 + Sin 60

13.

2 Sin (35º) Sin (15º)

Appendix

Here is three web sits to provide further information on Half Angle Identities

Double-Angle and Half-Angle Formulas

Half angle formulae

Hyperbolic Trigonometric Functions

Test

Questions and Answers:

Sin 165º

Sin (135º + 30º)

Sin 135º Cos 30º + Cos 135º Sin 30º

(√2)(√3) + (-√2)1

2 2 2 2

√6_ - √2_

4 4

√6 - √2

Tan (-15º)

Tan (30 – 45)

Tan 30 – Tan 45

1 + Tan 30 Tan 45

√3 - 1 √3 – 3

3 = 3___

1+ √3 3 + √3

3 3

√3 – 3

√3 + 3

1- Cos (2A)___ = Tan (A/2)

2 Sin A + Sin (2A)

___1 – (1- 2sin²A)___ = Tan (A/2)

2 Sin A + 2 SinA CosA

2 Sin²A____ = Tan (A/2)

2 SinA (1 + cosA)

Sin A__ = Tan (A/2)

1 + CosA

Tan (A/2) = Tan (A/2)

2 Sin²(A/2) + CosA = 1

2 (1 – CosA) + CosA = 1

1 = 1

Tan ½A + Cot ½A = 2 CscA

SinA + SinA = 2 CscA

1 + CosA 1 – CosA

SinA (1- CosA) + SinA ( 1 + CosA) = 2 CscA

1 – (Cos²A)

SinA – CosA SinA + SinA (1 + CosA) = 2 CscA

SinA²

2 SinA = 2 CscA

SinA²

2__ = 2 CscA

SinA

2 CscA = 2 CscA

SinB Tan (B/2) + 2 CosB = 2 Cos²(B/2)

SinB (1 – CosB) + 2CosB = 2 Cos²(B/2)

SinB

1 – CosB + 2 CosB = 2 Cos²(B/2)

1 + CosB = 2 Cos²(B/2)

2 Cos²(B/2) = 2 Cos²(B/2)

CosB – Tan(B/2) Sin(2B) – Cos(2B) = 2Sin²(B/2)

CosB – (1 – CosB) (2 SinB CosB) – ( Cos²B + Sin²B) = 2Sin²(B/2)

SinB

CosB – 2CosB + 2Cos²B - Cos²B + Sin²B = 2Sin²(B/2)

-CosB + Cos²B + Sin²B = 2Sin²(B/2)

-CosB + 1 = 2Sin²(B/2)

2(1 – CosB) = 2Sin²(B/2)

2Sin²(B/2) = 2Sin²(B/2)

Tan (B/2) = CscB – CotB

1 – CosB = CscB – CotB

SinB

1_ - CosB = CscB – CotB

SinB SinB

CscB – CotB = CscB – CotB

Cos80º Cos20º + Sin80ºSin20º

Cos (80-20)

Cos 60º

10.

Cos 2.4x – Cos 4.4x

-2 Sin (3.4x) Sin (-1x)

2 Sin (3.4x) Sin (1x)

11.

2 Cos (3.8x) Sin (4.1x)

Sin (7.9x) – Sin (-0.3x)

Sin (7.9x) + Sin (0.3x)

12.

Sin 30 + Sin 60

2 Sin (90/2) Cos (-30/2)

2 Sin (-15) Cos (-15)

2√2 Cos (15)

√2 Cos (15)

13.

2 Sin (35º) Sin (15º)

-(Cos (35-15) – Cos (35-15))

-(Cos (50º) – Cos (50º))