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

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).

Or perhaps discover a relationship for the angle sum less than π/2

[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]

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. ■

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)

[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]

, definition of the sine [4.01]

[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]

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))

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:

■

In figure 3, we note that [6.01]

Using the 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]

Trigonometry Contents

Ken Ward's Mathematics Pages

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