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Compound Angle Identities
Tutorial
Sin ( A + B ) = SinA CosB + CosA SinB
Sin ( A – B ) = SinA CosB – CosA SinB
Cos ( A + B) = CosA CosB – SinA SinB
Cos ( A – B ) = CosA CosB + SinA SinB
Tan ( A + B ) = TanA – TanB
1 - TanA TanB
Tan ( A – B ) = TanA – TanB
1 + TanA TanB
These are the identities that you can use along with
other identities to help solve problems.
Here are a few Web sites that will also help to explain
this topic…
 http://www.projectalevel.co.uk/maths/compound
.htm
 www.pinkmonkey.com/studyguides/trig/chap4/to
404301.htm
 http://www.acts.tinet.ie/compoundanglesandcalcu
_668.html
The Unit Circle, which is a very useful tool when
figuring out Compound Angle Identities looks like this:
Examples:
Sin(2A) = 2SinA CosA
Look at the formulas and change Sin (2A) to Sin (A + A)
Keep the Right side the same
Sin( A + A )= 2SinA CosA
Now you enter the formula for Sin (A + A) which is the same
as Sin (A + B)
SinA CosA + CosA SinA = 2SinA CosA
Then combine
2SinA CosA = 2SinA CosA
Example 2:
Cos (2A) = Cos2A – Sin2A
Again you change Cos (2A) to Cos (A+A) so that it is more
easily recognizable in order to finds the formula
Cos (A+A) = Cos2A – Sin2A
Plug in formula now
CosA CosA – SinA SinA = Cos2A – Sin2A
Combine
Cos2A – Sin2A= Cos2A – Sin2A
Example 3:
Tan (2A) = 2 Tan A
1-Tan2A
Tan ( A + A ) = 2 Tan A
1-Tan2A
TanA + Tan A = 2 Tan A
1- TanA TanA 1-Tan2A
2 Tan A = 2 Tan A
1-Tan2A
1-Tan2A
Now these are all questions that require the use of the
Unit Circle for determining values of original numbers.
In order to determine the proper numbers to use to
figure out the problems you must look at the angle that you
are trying to figure out and then look at what other angles can
add or subtract to give you that angle… ex. Sin 75 would
require the use of angles 45 and 30 because they add up to 75.
Therefore the formula that would be used in this case would
be (Sin (A + B)) which then ’provides’ the other formula that
solves the equation.
For Example to figure out Sin 75; ( example 1 (sine)):
Sin (75)
Sin ( 30 + 45 )
Now use the formula previously given
Sin30 Cos45 + Cos30 Sin45
The next step is to use the unit circle to help. You will notice
that Sin30 has another value, that is ½ and Sin45 also has
another value, that is 2/2 . Cos 30 is 3/2 and Cos 45 is also
2/2 . You then substitute the Sin and Cos values for their
‘square root’ values that we just listed from the Unit Circle.
Doing so then allows for further expanding and solving of
the problem…
½ 2/2
+ 3/3
2/2
Combine the equations
2/4
+
6/4
2 + 6
4
Example 2 ( Cosine ):
Cos ( 15 )
What equals 15 degrees… look at the unit circle…( 45 – 30 )
Cos ( 45 – 30 )
Plug in the formula using the given ( 45 and 30 ) degrees
numbers.
Cos45 Cos30 + Sin45 Sin30
2/2
3/2 + 2/2 1/2
Then combine the terms…
6/4
+
2/4
Make sure that the denominators are the same…
6 + 2
4
Example 3 ( Tangent ):
Tan ( -15 )
Look at unit circle just as before and figure out which angles
add up to give you the –15
Tan ( 30 – 45 )
=
Tan30 - Tan45
1 + Tan30 Tan45
=
3 – 1
1 + 3 + 1
=
(3 – 1 ) ( 1 - 3)
(1 + 3) (1 - 3)
* get rid of the root of the denominator ( by multiplying )*
=
23 – 4
-2
=
-3 +2
Now You Can Test Your Skills:
Compound Angle Identities Test:
(answers found in appendix)…
1.
2.
3.
4.
Sin(165)
Cos(80) Sin(20) + Cos(20) Sin(80)
Sin(37) Cos(7) - Cos(7) Sin(37)
Sin(A + B) = TanA + TanB
CosA CosB
5. Cos (45 – θ) = Sin (30 – θ)
6. Tan (A – θ)= 2/3
7. Cos2(26) – Sin2(26)
8. 2Cos2(34) – 1
9. 2Sin(14) Cos(14)
10. 1 – 2Sin24θ
11. Sinθ Cosθ = ½ Sin (2θ)
12. (SinA + CosA)(SinB + CosB)=Sin(A+B) +
Cos(A-B)
13. Sin ( A + B ) = TanA + TanB
CosA CosB
14. 1 – Cos (2B) + Sin2B = TanB
1 + Cos (2B) + Sin2B
15. 1 + TanX
1 - TanX
Appendix
Test Answers
1. Sin(165)
Sin(135+ 30)
Sin135 cos30 +cos135 sin30
√2/2 √3/2
-√2/2
½
√6/4 +
-√2/4
√6-√2
4
2. cos80 cos20 + sin80 sin20
cos(80-20)
cos60
½
3. sin37 cos7 – cos37 sin7
sin(37-7)
sin(30)
½
4. Sin(A + B) = TanA + TanB
CosA CosB
SinA CosB + CosA SinB = TanA + TanB
CosA CosB
SinA CosB + CosA SinB = TanA + TanB
CosACosB CosACosB
TanA + TanB = TanA TanB
5. Cos( 45 – θ) = Sin ( 30 – 0 )
Cos45 Cosθ + Sin45 Sinθ = Sin30 Cosθ + Cos30 Sinθ
√2/2 Cosθ + √2/2 Sinθ = ½ Cosθ + √3/2 Sin θ
√2/2 Cosθ – 1/2Cosθ = √3/2Sinθ - √2/2 Sinθ
√2 - 1 Cosθ = √3 - √2 Sin θ
2 Cosθ
2
Cosθ
2
√3- √2
√2 – 1 = √3 - √2 Tanθ
2
2
= √3 - √2
2
√2 – 1 = Tan θ
√3 - √2
6. Tan ( A – θ ) = 2/3
Tan (3 – θ) = 2/3
TanA – Tanθ = 2/3
1+TanA Tanθ
3 – Tanθ = 2/3
1+3Tanθ
9 -3 Tan θ = 2 + 6 Tan θ
-9tan θ = 7
Tan θ = -7/9
Tan θ = 0.7778
. θ = 37.9
7. Cos2 26 – sin2 26
Cos (26) cos (26) – Sin (26) sin (26)
Cos 52°
8. 2Cos2 (34)- 1
Cos 2 34 + Cos 2 34 - 1
Cos 2 34 – Sin 2 34
Cos 68°
9. 2Sin 14° Cos 14°
Sin 14 Cos 14 + Cos 14 Sin 14
Sin 28
10. 1-2 Sin2 4 θ
Cos (A+B)
Cos A Cos B – Sin A Sin B
11. Sin θ Cos θ = ½ Sin (2 θ)
Sin (2 θ)= 2 Sin θ Cos θ
12.(Sin A + Cos A) (Sin B + Cos B)= Sin (A+B) + Cos (A-B)
Sin A Sin B + Sin A Cos B + Cos A Cos B + Cos A Sin B
Cos A Cos B + Sin A Sin B + Sin A Cos B + Cos A Sin B
Cos (A+B)
Sin (A+B)
Sin (A+B) + Cos (A-B) = RHS
13. Sin (A+B) = tan A + tan B
Cos A Cos B
Sin A Cos B +Cos A Sin B = RHS
Cos A Cos B
Sin A Cos B + Cos A Sin B
Cos A Cos B
Cos A Cos B
Tan A + Tan B = RHS
14. 1-Cos (2B)+Sin 2B = tan B
1+ Cos (2B)+Sin 2 B
1- (Cos2 B – Sin 2 B) + 2 Sin B Cos B
1+(Cos2 B – Sin 2 B) + 2 Sin B Cos B
1-(1-2sin2B)+2SinB CosB
1+(2Cos 2 B –1) + 2SinB CosB
2 Sin2B + 2SinB CosB
2 Cos2B + 2SinB CosB
2SinB ( SinB+CosB)
2CosB ( CosB + CosB)
SinB = TanB
CosB
15. 1+TanX
1- TanX
Tan ( 45 + X ) = Tan45 + TanX
1 - (Tan45)(TanX)
Bibliography:
 Pre - Calculas Mathematics Two Pages 98 to 102
Websites:
 http://www.projectalevel.co.uk/maths/compound
.htm
 www.pinkmonkey.com/studyguides/trig/chap4/to
404301.htm
 http://www.acts.tinet.ie/compoundanglesandcalcu
_668.html
 (Our own notes) Mr. Paul Betuik