# Session Objectives

At the end of this session, candidates should be able to:
1) Use the addition formulae in solving simple trigonometric problems.

Consider the diagram above,

POX = A = QOP = B
QOX = POX + QOP
= A + B

Take a point C on OQ.
1)Draw a perpendicular from C to meet OP at D,

2) Draw a perpendicular from D to meet CF at G,

3) And lastly draw a perpendicular from D to meet OX at E

Note the element at the middle is equivalent to the angle e.g POX = 40° means O = 40° (I.e the triangle POX = 40°)

From the above diagram;
ODE = 90° – A

GDO = A since GDE = 90°

CDG = 90° – A since CDO = 90°,
hence GCD = A

Finding Expression For Sin(A + B), sin(A – B), cos(A + B), and cos(A – B) in terms of sinA, sinB, cosA and cosB

# First Expression

Find sin( A ± B)

From the above diagram,
Sin(A ± B) = sin QOX

= CF
OC
= CG + GF
OC
= CG+ GF
OC   OC
= GF + CG
OC    OC
= DE + CG
OC   OC
= DE × OD
OD   OC
= sinA × cosB
+ CG × CD
CD   OC
= cosA × sinB

hence;
Sin(A + B) = sinA.cosB + cosA.sinB
eqn (1)
Note : in mathematics . means multiplication

Substitute B to –B in equation (1)

Hence,

Sin(A – B) = sinA.cosB – cosA.sinB
eqn (2)

# Second Expression

Find cos(A ± B)

cos(A + B) =

= OF
OC
= OE – FE
OC
= OEFE
OC   OC
= OEGD
OC   OC
= OEOD
OD   OC
= OE × OD
OD   OC
= cosA × cosB
GD × CD
CD   OC
= sinA × sinB
cos(A + B) = cosA.cosB – sinA.sinB
Substitute B to –B

Hence

cos(A – B) = cosA.cos( –B) – sinA.sin( –B)
cos(A – B) = cosA.cosB + sinA.sinB

# Third Expression

Find tan(A ± B)
In mathematics tan

= Sin
Cos
tan(A + B)

= sin(A + B)
cos(A + B)
= sinAcosB + cosAsinB
cosAcosB – sinAsinB
Divide both numerator and denominator by cosAcosB

First numerator expression
tan (A + B)

=     sinAcosB +  cosAsinB
cosAcosB     cosAcosB
=   sinA + sinB
cosA    cosB
Denominator expression
tan (A + B)

=  cosAcosB –    sinAsinB
cosAcosB       cosAcosB
=  1 –   sinAsinB
cosAcosB
tan (A + B)

=    tanA + tanB
1 – tanAtanB
Substitute B to –B

tan (A – B)

=     tanA + tan( –B)
1 – tanAtan( –B)
Hence,

tan (A – B)

=    tanA – tanB
1 + tanAtanB
The formulae which we have expressed above are called Addition Formulae

note : Addition formulae are not only true for acute compound angles but are true for all compound angles.

Let look at some examples,

# Example 1

Using addition formulae, evaluate, sin 75°, sin 255°, cos 195°, cos 15°, and tan 195°.

# Solution 1

sin 75° = sin (30° + 45°)
= sin 30° + cos 45° + cos 30° + sin 45°
= 1 × √2 + √3 × √2
2
= ¼ × (√6 – √2)

# Solution 2

cos 15° = cos (45° – 30°)
= cos 45° cos 30 + sin 45° sin30°
= √2 × √3+ √2 × 1
2
= ¼ × (√6 + √2)

# Solution 3

sin 225° = sin (180° + 75°)
= sin 180° cos 75° + cos 180° sin 75°
= 0 – sin 75°
= – ¼ × (√6 + √2)

# Solution 4

cos 195° = cos (180° + 15°)
= cos 180°cos15° – sin 180°sin 15°
= –cos 15° – 0
= – ¼ × (√6 + √2)

# Solution 5

tan 195° = tan (180 + 15)

= tan 180° + 15°
1 – tan 180°tan15°
= 0 + tan15°
1 – 0
= tan 15°
3 – √3
3 + √3

# Common Angles And Their Equivalents

1) sin 60 = cos 30

= 0.866 or √3
2
2) sin30 = cos 60 = 0.5 or ½

3)

sin 45 = cos 45 = 0.7071 = 1
√2
4) tan 60 = 1.732 = √3

5)

tan 30 = 0.577 = 1
√3

# Summary

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
I think all those examples should do it, but if you still need more help don’t hesitate to ask, thank you.
Our next class topic is multiple angles.