Derive an identity that transforms sin(α+β+γ) into a sum of products of sines and/or cosines of the individual numbers α, β, γ.
sin(α+β+γ) , I will use sin(a+b+c) for easier typing
= sin( (a+b) +c)
= sin(a+b)cos c + cos(a+b)sin c
= (sinacosb + cosasinb)(cosc) + (cosacosb - sinasinb)sinc
= sinacosbcosc + cosacoscsinb + cosacosbsinc - sinasinbsinc
now put back your α,β, and γ
To derive the desired identity, we will make use of the trigonometric addition formula for sine:
sin(α + β) = sin α cos β + cos α sin β
Let's start by expanding sin(α + β + γ) using the addition formula twice:
sin(α + β + γ) = sin[(α + β) + γ]
= sin(α + β) cos γ + cos(α + β) sin γ
Now we can expand sin(α + β):
sin(α + β) = sin α cos β + cos α sin β
Substituting this into the expanded expression for sin(α + β + γ), we have:
sin(α + β + γ) = (sin α cos β + cos α sin β) cos γ + cos(α + β) sin γ
Since cos(α + β) = cos α cos β - sin α sin β, we can further simplify:
sin(α + β + γ) = (sin α cos β + cos α sin β) cos γ + (cos α cos β - sin α sin β) sin γ
Expanding this expression, we get:
sin(α + β + γ) = sin α cos β cos γ + cos α sin β cos γ + cos α cos β sin γ - sin α sin β sin γ
Now we notice that sin β cos γ can be simplified using the addition formula for sine:
sin(β + γ) = sin β cos γ + cos β sin γ
Therefore, we can substitute this expression in the previous equation:
sin(α + β + γ) = sin α cos β cos γ + cos α (sin β cos γ + cos β sin γ) + cos α cos β sin γ - sin α sin β sin γ
Rearranging terms, we obtain the desired identity:
sin(α + β + γ) = sin α cos β cos γ + cos α sin β cos γ + cos α cos β sin γ - sin α sin β sin γ
This identity transforms sin(α + β + γ) into a sum of products of sines and/or cosines of the individual numbers α, β, γ.
To derive an identity that transforms sin(α+β+γ) into a sum of products of sines and/or cosines of the individual numbers α, β, γ, we can use the angle addition formula. The angle addition formula states that:
sin(x + y) = sin(x)cos(y) + cos(x)sin(y)
Using this formula, we can expand sin(α+β+γ) as follows:
sin(α+β+γ)
= sin((α+β)+γ)
= sin(α+β)cos(γ) + cos(α+β)sin(γ)
Next, we can apply the angle addition formula again to sin(α+β) and cos(α+β):
sin(α+β)cos(γ) + cos(α+β)sin(γ)
= (sin(α)cos(β) + cos(α)sin(β))cos(γ) + (cos(α)cos(β) - sin(α)sin(β))sin(γ)
Expanding further, we get:
= sin(α)cos(β)cos(γ) + cos(α)sin(β)cos(γ) + cos(α)cos(β)sin(γ) - sin(α)sin(β)sin(γ)
Now, we can rearrange the terms to obtain the desired form:
= sin(α)cos(β)cos(γ) + sin(β)cos(α)cos(γ) + sin(γ)cos(α)cos(β) - sin(α)sin(β)sin(γ)
Therefore, the derived identity is:
sin(α+β+γ) = sin(α)cos(β)cos(γ) + sin(β)cos(α)cos(γ) + sin(γ)cos(α)cos(β) - sin(α)sin(β)sin(γ)
This identity transforms sin(α+β+γ) into a sum of products of sines and cosines of the individual numbers α, β, γ.