The Pythagorean triganometric identity
sin2θ+cos2θ=1Proof:
In this right triangle …
Pythagoras’ Theorem states that a2+b2=c2. Dividing both sides by c2 then gives us a2c2+b2c2=1. Since the definitions of sinθ and cosθ are, respectively, bc and ac, the above identity follows immediately.
QED.
Angle sum and difference identities
sin(α±β)=sinαcosβ±cosαsinβ cos(α±β)=cosαcosβ∓sinαsinβ tan(α±β)=tanα±tanβ1∓tanαtanβProof:
Consider the following …
Here …
sinα=BCOC=DCPC sinβ=PCOP cosα=OBOC=DPPC cosβ=OCOP sin(α+β)=APOP=AD+DPOP=BC+DPOP=BCOP+DPOP=BCOCOCOP+DPPCPCOP=sinαcosβ+cosαsinβ cos(α+β)=OAOP=OB−ABOP=OB−DCOP=OBOP−DCOP=OBOCOCOP−DCPCPCOP=cosαcosβ−sinαsinβIf we substitute −β for β in the above then we get …
sin(α+−β)=sinαcos(−β)+cosαsin(−β) cos(α+−β)=cosαcos(−β)−sinαsin(−β)But we know that cos(−β)=cosβ and sin(−β)=−sinβ and so …
sin(α−β)=sinαcosβ−cosαsinβ cos(α−β)=cosαcosβ+sinαsinβNow, since …
tanα=sinαcosαThen …
tan(α+β)=sin(α+β)cos(α+β)=sinαcosβ+cosαsinβcosαcosβ−sinαsinβIf we divide the top and bottom by cosαcosβ then we get …
=sinαcosβcosαcosβ+cosαsinβcosαcosβcosαcosβcosαcosβ−sinαsinβcosαcosβ =sinαcosα+sinβcosβ1−sinαsinβcosαcosβ =tanα+tanβ1−tanαtanβIf we substitute −β for β in the above then we get …
tan(α+−β)=tanα+tan(−β)1−tanαtan(−β)But we know that tan(−β)=−tanβ and so …
tan(α−β)=tanα−tanβ1+tanαtanβQED.