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Compound Angles: Geometrical Proof for sin(A+B)


    what you'll learn...

sin(A+B)

sin a+b proof  »  the coordinates of points
    →  P(cosA,sinA)
    →  Q(cosB,sinB)
    →  R(cos(A+B),sin(A+B))
    →  Q1(sinB,cosB)
    →  T(0,1)

 »  equate the distance RT¯=PQ1¯

 »  sin(A+B)=sinAcosB+cosAsinB


cos(A+B)

cos a+b proof  »  equate the distance RS¯=PQ2¯

 »  cos(A+B)=cosAcosB-sinAsinB

problem definition

Consider points P and Q with angles A and B in unit circle as shown in the figure.

The point P is (cosA,sinA)

The point Q is (cosB,sinB)

sine A+B proof

In the given figure, point R is for angle (A+B).

The point R is (cos(A+B),sin(A+B)).

In the given figure, coordinates of P, and Q are given. Can coordinate of R be calculated?

In other words, given the trigonometric ratios of A and B , can we compute trigonometric ratios of A+B?
sin(A+B)=?
cos(A+B)=?

proof

sine A+B proof

In the modified figure, point Q1 is for angle (90-B).

The point Q1 is (cos(90-B),sin(90-B)) =(sinB,cosB) As cos(90-B)=sinB and sin(90-B)=cosB.

In the given figure, consider point T(0,1).

the angle ROT

    =SOT-SOR

    =90-(A+B)

In the given figure,

POQ1
    =SOT-SOP-Q1OT
    =90-A-B

In the given figure:
ROT=90-(A+B)
POQ1=90-A-B.

That is, the angle subtended by the two chords RT¯ and PQ1¯ are equal.

In the given figure, the coordinate of R is (cos(A+B),sin(A+B)) and the coordinate of T is (0,1). So, the Square of length of chord, using the distance formula, is

RT¯2

    =(cos(A+B)-0)2+(sin(A+B)-1)2

    =2-2sin(A+B)

Square of length of chord RT¯ is calculated as follows.

T(0,1)
R(cos(A+B),sin(A+B))

RT¯2
    =(cos(A+B)-0)2
      +(sin(A+B)-1)2
    =cos2(A+B)+sin2(A+B)
      +1-2sin(A+B)
    =2-2sin(A+B)

In the given figure, the coordinate of P is (cosA,sinA). and the coordinate of Q1 is (sinB,cosB).

The square of length of chord is

PQ1¯2

    =(cosA-sinB)2+(sinA-cosB)2

    =2-2cosAsinB-2sinAcosB

Square of length of chord PQ1¯ is calculated as follows.

P(cosA,sinA)
Q1(sinB,cosB)

PQ1¯2
    =(cosA-sinB)2
      +(sinA-cosB)2
    =cos2(A)+sin2(A)
      +sin2(B)+cos2(B)
      -2cosAsinB-2sinAcosB
    =2-2cosAsinB-2sinAcosB

Equating the square of lengths of chords RT¯ and PQ1¯.
2-2sin(A+B)=2-2cosAsinB-2sinAcosB

This proves sin(A+B)=sinAcosB+cosAsinB

proof for cosine

proof for cos(A+B)

To compute the value of cos(A+B) the enclosed figure is used and the proof is outlined below.

P(cosA,sinA)
Q(cosB,sinB)
R(cos(A+B),sin(A+B))
S(1,0)
Q2(cosB,-sinB)

PQ2¯2
    =(cosA-cosB)2+(sinA+sinB)2
    =2-2cosAcosB+2sinAsinB

RS¯2
    =(cos(A+B)-1)2+(sin(A+B)-0)2
    =2-2cos(A+B)

equating these two cos(A+B)=cosAcosB-sinAsinB

summary

sin(A+B)=sinAcosB+cosAsinB
cos(A+B)=cosAcosB-sinAsinB

Outline

It is advised to do the firmfunda version of "basics of Trigonometry" course before doing this.

The outline of material to learn "Advanced Trigonometry" is as follows.
Note: go to detailed outline of Advanced Trigonometry

    →   Unit Circle form of Trigonmetric Values

    →   Trigonometric Values in all Quadrants

    →   Trigonometric Values or any Angles : First Principles

    →   Understanding Trigonometric Values in First Quadrant

    →   Trigonometric Values in First Quadrant

    →   Trigonometric Values of Compound Angles: Geometrical Proof

    →   Trigonometric Values of Compound Angles: Algebraic Proof

    →   Trigonometric Values of Compound Angles: tan cot

    →   Trigonometric Values of Compound Angles: more results