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°
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:
Or perhaps discover a relationship for the angle sum less than
π/2
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.]
, definition of the
sine [4.01]
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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