Triganometric identities - Proofs

By geometry

The Pythagorean triganometric identity

sin2θ+cos2θ=1

Proof:

In this right triangle …

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)=cosαcosβsinαsinβ tan(α±β)=tanα±tanβ1tanαtanβ

Proof:

Consider the following …

Locked - Unlocked - Locked

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=OBABOP=OBDCOP=OBOPDCOP=OBOCOCOPDCPCPCOP=cosαcosβsinαsinβ

If 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(αβ)=sinαcosβcosαsinβ cos(αβ)=cosαcosβ+sinαsinβ

Now, since …

tanα=sinαcosα

Then …

tan(α+β)=sin(α+β)cos(α+β)=sinαcosβ+cosαsinβcosαcosβsinαsinβ

If we divide the top and bottom by cosαcosβ then we get …

=sinαcosβcosαcosβ+cosαsinβcosαcosβcosαcosβcosαcosβsinαsinβcosαcosβ =sinαcosα+sinβcosβ1sinαsinβcosαcosβ =tanα+tanβ1tanαtanβ

If we substitute β for β in the above then we get …

tan(α+β)=tanα+tan(β)1tanαtan(β)

But we know that tan(β)=tanβ and so …

tan(αβ)=tanαtanβ1+tanαtanβ

QED.