

geometric proofs of trigonometric identities 


Introduction
Below, some proofs are presented of trigonometric identities.
The proofs do not use any trigonometric formula or rule, what makes them quite special.
Identity Nr. 1
arctan( ) +
+ arctan( ) =
arctan(1 )
Proof A.
See figure 1. below.
Pictured are 4 squares.

fig.1 
Since:
LC_{1} = arctan( )
LD_{1} = arctan( )
LB_{1} = arctan(1 )
We must prove:
LC_{1} + LD_{1} = LB_{1}.........1)
DEMC ~ DDAE, because:
1. LA = LM = 90^{0}
2. MC = 3.ME
3. AD = 3.AE
so:
which changes ........1) into:
This is a basic geometric theorem.
(remember: 180^{0}  LB_{1} = 180^{0}  (LC_{1}+LE_{1})
This concludes proof A.
Proof B.
See figure 2. below:

fig.2 
Note: EF = 5, because of theorem of Pythagoras.
DADG is the rotation of DABE over 90^{0}= LGAE.
Now observe polygon AEFG.
so:
DAFG and DAFE are congruent
so:
LGAF = LFAE = 45^{0}
LGAD + LDAF = 45^{0}
arctan + arctan = arctan 1
Which concludes proof B.
Identity Nr. 2
sin(20^{0} ) + sin(40^{0} ) = sin(80^{0} )
Proof.
See figure 3. below:

fig.3 
M is the center of a circle with radius = 1.
Points A,B,C,D are located on the circle at the indicated angles.
Chord DC is extended by CE = BC.
Now DCBE is equilateral because:
LDCB = 0.5(360^{0}  120^{0}) = 120^{0 }.........(see appendix 1.)
LECB = 60^{0} ....so
LCBE = LCEB = 60^{0}....so
BE = BC = AB
and we see:
DDBE is congruent to DDBA....so
DE = DA ............and since
CE + DC = DE
AB + DC = AD
2sin(20^{0} )+ 2sin(40^{0} ) = 2sin(80^{0} )........(see appendix 2)
sin(20^{0} )+ sin(40^{0} ) = sin(80^{0} )
and we have proved the 2nd identity.
Appendix 1 (measuring angles by arcs)
See figure 4. below:

fig.4 
By:...... arc AB = 80^{0} we mean : LAMB = 80^{0}
Since
LM_{1} = LA + LP_{1}.......exterior angle of DPMA
MA = MP
LP_{1}= LA
LM_{1} = 2*LP_{1}
LP_{1} = 0.5*LM_{1} = 0.5* arc AC.....2
.....1 and .....2 combined:
LP_{1,2} = 0.5*LM_{1,2} = 0.5*arc AB
In words:
an angle located on a circle is half the size of the associated arc.
Appendix 2 (sine and chord relation)
See figure 5. below:

fig.5 

