Notes to a video lecture on http://www.unizor.com

Conditional Trigonometric Identities

Regular trigonometric identities represent equalities of certain trigonometric functions of angles without any restrictions on the angles.
For example,
sin(φ) = cos(π/2−φ)
or
sin(α+β) =
= sin(α)·cos(β)+cos(α)·sin(β).

Conditional trigonometric identities are true only if certain condition about the angles involved is true.
Below is an example of such conditional identity with the proof.

Problem
If α, β and γ are angles of some triangle then
sin(2α)+sin(2β)+sin(2γ) =
= 4·sin(α)·sin(β)·sin(γ)

Proof
Let's try "brute force" method by invariantly transforming the left side of this identity to the right sides using trigonometric functions of single angles and taking into account given condition on participating angles.

Since α, β and γ are angles of a triangle,
α+β+γ=π
and, therefore,
γ=π−(α+β).
Hence,
sin(γ)=sin(π−(α+β))=sin(α+β)
and
sin(2γ) = sin(2π−2α−2β) =
= sin(−2α−2β) = −sin(2α+2β)
.

Let's express sin of double angles in terms of trigonometric functions of single angles and also use the last identities for angle γ in terms of α and β to invariantly transform the left side of our original identity.
sin(2α)+sin(2β)+sin(2γ) =
=sin(2α)+sin(2β)−sin(2α+2β)=
= 2sin(α)·cos(α)+
+ 2sin(β)·cos(β)−
− sin(2α)·cos(2β)−
− cos(2α)·sin(2β) =
= 2sin(α)·cos(α)+
+ 2sin(β)·cos(β)−
− 2sin(α)·cos(α)·[1−2sin²(β)]− − 2sin(β)·cos(β)·[1−2sin²(α)] =
= 2sin(α)·cos(α)+
+ 2sin(β)·cos(β)−
− 2sin(α)·cos(α)+
+ 2sin(α)·cos(α)·2sin²(β)−
− 2sin(β)·cos(β)+
+ 2sin(β)·cos(β)·2sin²(α) =
= 4sin(α)·cos(α)·sin²(β)+
+ 4sin(β)·cos(β)·sin²(α) =
= 4sin(α)·sin(β)·
·[cos(α)·sin(β)+sin(α)·cos(β)] =
= 4sin(α)·sin(β)·sin(α+β) =
= 4sin(α)·sin(β)·sin(γ)
End of proof.

Here is another example.

Problem
If α = β + γ then
sin²(α)+sin²(β)+sin²(γ) =
= 2·[1−cos(α)·cos(β)·cos(γ)]

Proof
Obviously, the first intention is to express sin and cos of angle α on both sides of an identity in terms of angles β and γ, substitute it in the identity above and see if we get an unconditional identity.
sin(α)=sin(β)cos(γ)+cos(β)sin(γ)
sin²(α)=
= sin²(β)cos²(γ)+cos²(β)sin²(γ)+
+ 2sin(β)cos(γ)cos(β)sin(γ)
cos(α)=cos(β)cos(γ)−sin(β)sin(γ)

Now we can transform the original identity to the invariant one without angle α and, therefore, unconditional:
sin²(β)cos²(γ)+cos²(β)sin²(γ)+
+ 2sin(β)cos(γ)cos(β)sin(γ)+
+ sin²(β)+sin²(γ) =
= 2−2cos²(β)cos²(γ)+
+ 2sin(β)cos(β)sin(γ)cos(γ)

Both left and right parts contain the same component
2sin(β)cos(β)sin(γ)cos(γ)
that we can safely drop from the identity.
Therefore, we have to prove that
sin²(β)cos²(γ)+cos²(β)sin²(γ)+
+ sin²(β)+sin²(γ) =
= 2−2cos²(β)cos²(γ)

Replacing sin²() with 1−cos²() on the left side, we obtain
[1−cos²(β)]cos²(γ)+
+ cos²(β)[1−cos²(γ)]+
+ sin²(β)+sin²(γ) =
= cos²(γ)−cos²(β)cos²(γ)+
+ cos²(β)−cos²(β)cos²(γ)+
+ sin²(β)+sin²(γ) =
= sin²(β)+cos²(β)+
+ sin²(γ)+cos²(γ)−
−2cos²(β)cos²(γ)=
= 2−2cos²(β)cos²(γ)=
= 2·[1−cos²(β)cos²(γ)]
End of proof.