nav.gif (1245 bytes)mathematics

Ken Ward's Mathematics Pages

Trigonometry - Compound Angles

Trigonometry Contents

Page Contents

On this page, we claim to prove the sine and cosine relations of compound angles in a triangle, considering the cases where the sum of the angles is less than or more than 90°, and when one of the angles is greater than 90°
  1. Angle (α+β)<π/2
    1. Proof of the Sine and Cosine Compound Angles
      1. Proof of sin(α+β)=sinα cosβ +cosα sineβ
      2. Proof of cos(α+β)=sinα cosβ −cosα sineβ
  2. Angle (α+β)>π/2
    1. Proof of sin(α+β)=sinα cosβ +cosα sineβ, when α+β>π/2
    2. Proof of cos(α+β)=cosα cosβ +sinα sineβ, when α+β>π/2
  3. Angle (α+β)>π/2, and β>π/2
    1. Proof of sin(α+β)=sinα cosβ +cosα sinβ, when α+β>π/2, and β>π/2
    2. Proof of cos(α+β)=cosα cosβ − sinα sinβ, when α+β>π/2, and β>π/2

Angle (α+β)<π/2

The diagram below is composed of two right angled triangles, (OAB and OAE). The angle EOA is called α, and the angle AOB,  β.  The sum , α+β<90°. BD is perpendicular to OE and AX is perpendicular to BD, meeting it at X.
We note the following angles in the diagram (using "∠" for angle):
∠XAO=α (EOA and XAO are alterate angles of parallel lines, XA and OE)
∠XAB = 90- (OAB is a right angle, and XAB is 90°-α)
So angle ABX=α (triangle XAB is a right angled triangle, so ABX=90°-(90°-α)=α)
Also, ∠OBA in triangle ABO is (90°-β), so angle OAB=90°-α-β
Angles of interest are marked on the diagram (Fig 1).

Proof of the Sine and Cosine Compound Angles

Proof of sin(α+β)=sinα cosβ +cosα sineβ

We wish to prove that:
sin A+B
Or perhaps discover a relationship for the angle sum less than π/2
From the diagram above, we note:
sinab1[2.0], by the definition of sine.
We wish to obtain an expression for BD, and note that:
sinab2[2.01] (because both are segments of the line BD)

From triangles XAB, and OAE, BX=BA·cosα, and XD=OA·sinα, and substituting these values in 2.01, we obtain:
sinab3 [2.02]
Dividing this (2.02) by OB, we obtain (because sin(α+β)=BD/OB, equation 2.0)
sinab4 [2.03]
By definition of sine and cosine and from triangle OAB, we obtain:
sinab5 [2.04]
Substituting these values in 2.03, we obtain the required relationship:
sinab6 [2.05]

Proof of cos(α+β)=sinα cosβ −cosα sineβ

(Fig 1 is repeated below. See above for the explanation of the angles)compoundAngle1
From the triangle OBD, we note that for angle OBD:
Recalling that sin(90-A)=cos(A):
cosab1 [2.07]
cosab3 [2.08]
From the diagram we note:
cosab4 (Because OE is a straight line) [2.09]

From ΔOAB, we find:
cosagb (From the definitions of sine and cosine) [2.10]
Substituting this value for OD in equation 2.08, we have:
cosab6 [2.11]
From ΔOAB:
cosab7  (Definitions of sine and cosine) [2.12]
And substituting this in equation 2.11:
cosab8  [2.13] 
Which is the cosine of the sum of two angles in the acute angled triangle ODB.

Angle (α+β)>π/2

In Fig 2, ∠EOA=α and ∠AOB=β, and the sum α+β is greater than 90°.
BA is perpendicular to OA and the line BXD is perpendicular to the line DOE.
Line AX is perpendicular to BD.
Because angle OAB in ΔOAB is a right angle, then angle OBA=90°−β (as marked in the diagram)
In ΔODB, ∠DOB=180°−(α+β) (Angle of a straight line is 180°). And ∠DBO=90°-(180°−(α+β) )=(α+β)−90° (as marked in the diagram).
So, angle DBA=α (the sum of angles DBO and OBA)
Compound Angle 2

Proof of sin(α+β)=sinα cosβ +cosα sineβ, when α+β>π/2

In Fig 2, we note, by the definition of cosine:
sin22ab1 [3.01]
Because for any angle θ, cos(θ-π/2)=cos(-(π/2-θ)=cosθ, equation 3.01 becomes:
sin22ab2 [3.02]
sin22ab3 [3.03]
From equation 3.01, we obtain:
sin22ab6 (Segments of the line BD[3.06]
Using the definitions of sine and cosine in triangles ABX and OBD:
sin22ab7 [3.07]
Substituting BD in equation 3.05
sin22ab8 [3.08]
From triangle OAB and using the relationship, cos(π/2-β)=sin(β), we find:
sin22ab9 [3.09]
sin22ab10 [3.10]
Substituting these values in 3.08 and simplifying, we find:
sin22ab11 [3.11]
By starting the values with α (for presentation):
sin22ab12 ■ [3.12]

Proof of cos(α+β)=cosα cosβ +sinα sineβ, when α+β>π/2

[Fig 2 is repeated below for convenience.]
 Compound Angle 2

sin2ab2, definition of the sine [4.01]
Also from ΔDBO, we note for the angle DBO
Because sin(π/2-θ)=cosθ, and because the cosine of a negative angle is positive.
sin2ab3, from 4.02 and 4.01 [4,03]
From the diagram we can find DO
sin2ab5, segments of the line DE [4.04]
From the definitions of the sine and cosine in ΔDBO ΔOBA
sin2ab4 [4.05]
Substituting in equation 4.03, we find:
sin2ab7 [4.06]

We also note that in ΔOBA, the following relationships hold:
sin2ab6, definition of sine and cosine [4.07]
Substituting these values in 4.06
sin2ab8 [4.07]
Expanding 4.07, we have:
sin2ab9 [4,08]
And rearranging to put alpha in the first positions (for cosmetic reasons only), we have
sin2ab10  [4.09]

Angle (α+β)>π/2, and β>π/2

In Fig 3, below, angle POF=α, and angle FOB =β
BX and EA are parallel and perpendicular to DEO, which is parallel to XA.
BA is a line drawn perpendicular to FA from B
The angle AOD=α (Opposite angles). OAE=(π/2-α) (Angle sum of ΔOAE), so ∠BAE=α because it equals π/2−(π/2-α)
Angle OQB=(π/2+α) because it is the external angle of ΔAEQ, and is the sum of QEA (π/2) and EAQ(α).
In triangle OQB, the external angle POB (α+β)=π/2+α +∠QBO, hence, ∠QBO=β−π/2
∠DBQ=the external angle OQB (π/2+α)−QDB (π/2)=α
Finally, we note that ∠OBD=α+β−π/2
cos(α+β−π/2)=cos(−(π/2−(α+β))=cos((π/2−(α+β)) =sin(((α+β))
sin(α+β−π/2)=sin(−(π/2−(α+β))=−sin((π/2−(α+β)) =−cos(((α+β))


Proof of sin(α+β)=sinα cosβ +cosα sinβ, when α+β>π/2, and β>π/2

In Fig 3, we note that:sin331 [5.01]
And the cosine is:
sin332 [5.02]
Rearranging the cosine to the form cos(π/2-(α+β)
sin333 [5.03]
Hence, from 5.02, we have
sin334 [5.04]
sin335 [5.05]
We have from ΔABD, using the definitions of sine and cosine:
sin336 [5.06]
Dividing by BO throughout, and noting that DB/BO=sin(α+β) from 5.02
sin337 [5.07]
That is, (omitting the first equality)
sin338 [5.08]
From ΔABO, and the definitions of sine and cosine, and the formula for complementary angles for sines and cosines[sin(π/2-θ)=cos(θ), cos(π/2-θ)=sin(θ)]
sin339 [5.09]
sin3310 [5.10]
Substituting the values for AB/BO and AO/BO in 5.08, we obtain the formula for the sine of a compound angle, when one is greater than π/2
sin3311 [5.11]
Which, by putting sin(α) first, for cosmetic reasons:

Proof of cos(α+β)=cosα cosβ − sinα sinβ, when α+β>π/2, and β>π/2

Figure 3 is repeated below. The angles are founds as before. compoundAngle3
In figure 3, we note that cos331 [6.01]
Using the relationship between the sines and cosines of complementary angles:
cos332 [6.02]
Because for the general angle θ, sin(θ-π/2)=sin[-(π/2−θ)]=−sin(π/2−θ)=−cos(θ)

Hence, in ΔBDO
We note for the line DEO=DE+EO, and because XA=DE, we have:
cos334 [6.04]
From ΔABX and ΔAOE and the definitions of cosine and sine
cos335 [6.05]
Dividing throughout by DO:
cos336 [6.06]
From 6.03, we have:
cos337 [6.07]
Hence, substituting this in 6.06
cos337b [6.07b]
cos338 [6.08]
And also from ΔABO
cos339 [6.09]
Substituting these values in 6.07b:
cos3310 [6.10]
Multiplying throughout by minus one, and slightly rearranging:
cos3311 [6.11]

Trigonometry Contents

Ken Ward's Mathematics Pages

Faster Arithmetic - by Ken Ward

Ken's book is packed with examples and explanations that enable you to discover more than 150 techniques to speed up your arithmetic and increase your understanding of numbers. Paperback and Kindle:
Buy now at
Faster Arithmetic for Adults