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
Cos
5. Cot = 1
Tan
6. Sec = 1
Cos
7. Csc = 1
Sin
By using the double angle equations 3 and 4 we may derive the equations for the halfangles.
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)
2
Since we are finding the HALF-angles we must find HALF of A. To do this you must
replace every A with A:
2
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:
Replace A with A:
2
Sin2A = 1 - Cos(2A)
2
Sin2(A) = 1 - Cos(2A/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 halfangle 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)
Cos2(A/2)
which equals
Tan2(A/2) = 1 - CosA
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
SinA
Half-angle identity 6: TanA/2 = 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
2
And: (A+B) - (A-B) = C-D
Which equals: B = C-D
2
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)
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
sinx
= 1-cosx +2cosx
= 1+cosx
= 2Cos2(x/2)
substitute for tan half-angle
expand and simplify
rearrange cosine half angle Identity
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
2
2
=
2cos 5x cos x
2
2
Cast Rule:
Only Sine is positive
in this quadrant.
All (Sine, Cosine, Tangent, etc) are
positive in this quadrant.
IV
S
A
І
III
T
C
II
Only Tangent is positive
in this quadrant.
Only Cosine is positive in this
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
d) cosx - cos3x
2
2
b) cos2x + cos 4x
e) sin6x + sin2x
c) sinx + sin3x
f) cos5x - cos 3x
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.
Sin 165º
2.
Tan (-15º)
Part II
Prove the following equations that they are identities.
3.
1- Cos (2A)___ = Tan (A/2)
2 Sin A + Sin (2A)
5.
Tan ½A + Cot ½A = 2 CscA SinB
4.
2 Sin²(A/2) + CosA = 1
6.
Tan (B/2) + 2 CosB = 2 Cos²(B/2)
7.
CosB – Tan(B/2) Sin(2B) – Cos(2B) = 2Sin²(B/2)
8.
Tan (B/2) = CscB – CotB
Part III
Transform the indicated Sum (or Difference) into a product of sins and cosines of
positive arguments.
9.
Cos80º Cos20º + Sin80ºSin20º
10.
Cos 2.4x – Cos 4.4x
11.
2 Cos (3.8x) Sin (4.1x)
12.
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:
1.
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
4
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
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)
4.
2 Sin²(A/2) + CosA = 1
2 (1 – CosA) + CosA = 1
2
1=1
5.
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
6.
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)
7.
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)
2
2Sin²(B/2) = 2Sin²(B/2)
8.
Tan (B/2) = CscB – CotB
1 – CosB = CscB – CotB
SinB
1_ - CosB = CscB – CotB
SinB
SinB
CscB – CotB = CscB – CotB
9.
Cos80º Cos20º + Sin80ºSin20º
Cos (80-20)
Cos 60º
½
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
√2 Cos (15)
13.
2 Sin (35º) Sin (15º)
-(Cos (35-15) – Cos (35-15))
-(Cos (50º) – Cos (50º))