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Trigonometry - Compound Angles

Trigonometry Contents

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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)
From the diagram above, we note:
[2.0], by the definition of sine.
We wish to obtain an expression for BD, and note that:
[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:
[2.02]
Dividing this (2.02) by OB, we obtain (because sin(α+β)=BD/OB, equation 2.0)
[2.03]
By definition of sine and cosine and from triangle OAB, we obtain:
[2.04]
Substituting these values in 2.03, we obtain the required relationship:
■ [2.05]

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

(Fig 1 is repeated below. See above for the explanation of the angles)
From the triangle OBD, we note that for angle OBD:
[2.06]
Recalling that sin(90-A)=cos(A):
[2.07]
Hence,
[2.08]
From the diagram we note:
(Because OE is a straight line) [2.09]


From ΔOAB, we find:
(From the definitions of sine and cosine) [2.10]
Substituting this value for OD in equation 2.08, we have:
[2.11]
From ΔOAB:
 (Definitions of sine and cosine) [2.12]
And substituting this in equation 2.11:
 [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)

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

In Fig 2, we note, by the definition of cosine:
[3.01]
Because for any angle θ, cos(θ-π/2)=cos(-(π/2-θ)=cosθ, equation 3.01 becomes:
[3.02]
[3.03]
[3.04]
From equation 3.01, we obtain:
[3.05]
Now,
(Segments of the line BD[3.06]
Using the definitions of sine and cosine in triangles ABX and OBD:
[3.07]
Substituting BD in equation 3.05
[3.08]
From triangle OAB and using the relationship, cos(π/2-β)=sin(β), we find:
[3.09]
And
[3.10]
Substituting these values in 3.08 and simplifying, we find:
[3.11]
By starting the values with α (for presentation):
■ [3.12]

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

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

We also note that in ΔOBA, the following relationships hold:
, definition of sine and cosine [4.07]
Substituting these values in 4.06
[4.07]
Expanding 4.07, we have:
[4,08]
And rearranging to put alpha in the first positions (for cosmetic reasons only), we have
■  [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: [5.01]
And the cosine is:
[5.02]
Rearranging the cosine to the form cos(π/2-(α+β)
[5.03]
Hence, from 5.02, we have
[5.04]
Because
[5.05]
We have from ΔABD, using the definitions of sine and cosine:
[5.06]
Dividing by BO throughout, and noting that DB/BO=sin(α+β) from 5.02
[5.07]
That is, (omitting the first equality)
[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)]
[5.09]
[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
[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.
In figure 3, we note that [6.01]
Using the relationship between the sines and cosines of complementary angles:
[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
[6.03]
We note for the line DEO=DE+EO, and because XA=DE, we have:
[6.04]
From ΔABX and ΔAOE and the definitions of cosine and sine
[6.05]
Dividing throughout by DO:
[6.06]
From 6.03, we have:
[6.07]
Hence, substituting this in 6.06
[6.07b]
From ΔABO
[6.08]
And also from ΔABO
[6.09]
Substituting these values in 6.07b:
[6.10]
Multiplying throughout by minus one, and slightly rearranging:
■ [6.11]

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Faster Arithmetic - by Ken Ward