Derivation of Sum and Difference of Two Angles

Triangle used in sum and difference of two anglesThe sum and difference of two angles can be derived from the figure shown below.
 

Consider triangle AEF:
cosβ=¯AE1;¯AE=cosβ

sinβ=¯EF1;¯EF=sinβ
 

From triangle EDF:
sinα=¯DE¯EF

sinα=¯DEsinβ

¯DE=sinαsinβ
 

cosα=¯DF¯EF

cosα=¯DFsinβ

¯DF=cosαsinβ
 

¯BC=¯DE=sinαsinβ
 

From Triangle ACE:
sinα=¯CE¯AE

sinα=¯CEcosβ

¯CE=sinαcosβ
 

cosα=¯AC¯AE

cosα=¯ACcosβ

¯AC=cosαcosβ
 

¯BD=¯CE=sinαcosβ
 

The summary of the above solution is shown below:
 

summary-sum-and-difference-of-two-angles.jpg

 

Sum of two angles
From triangle ABF:
sin(α+β)=¯BD+¯DF

sin(α+β)=sinαcosβ+cosαsinβ

 

cos(α+β)=¯AC¯BC

cos(α+β)=cosαcosβsinαsinβ

 

tan(α+β)=sin(α+β)cos(α+β)

tan(α+β)=sinαcosβ+cosαsinβcosαcosβsinαsinβ

tan(α+β)=sinαcosβcosαcosβ+cosαsinβcosαcosβcosαcosβcosαcosβsinαsinβcosαcosβ

tan(α+β)=sinαcosα+sinβcosβ1sinαcosαsinβcosβ

tan(α+β)=tanα+tanβ1tanαtanβ

 

Difference of two angles
Let β = -β and note that
sin (-β) = -sin β
cos (-β) = cos β and
tan (-β) = -tan β
 

sin[α+(β)]=sinαcos(β)+cosαsin(β)

sin(αβ)=sinαcosβcosαsinβ

 

cos[α+(β)]=cosαcos(β)sinαsin(β)

cos(αβ)=cosαcosβ+sinαsinβ

 

tan[α+(β)]=tanα+tan(β)1tanαtan(β)

tan(αβ)=tanαtanβ1+tanαtanβ