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Why is sinα−sinβ=2cos(α+β2)sin(α−β2)sin⁡α−sin⁡β=2cos⁡(α+β2)sin⁡(α−β2)?
2 Answers

Daniel McLaury, Ph.D. Student in Mathematics at University of Illinois at Chicago
419 Views · Daniel has 810+ answers and 21 endorsements in Mathematics
Clearly having a formula for sinα−sinβsin⁡α−sin⁡β would be nice, so if you saw something that looked like it could lead there then you'd probably try to make it work out.

Of course you already know the angle-sum identity for sine,

sin(θ+φ)=sin(θ)cos(φ)+sin(φ)cos(θ)sin⁡(θ+φ)=sin⁡(θ)cos⁡(φ)+sin⁡(φ)cos⁡(θ)

Given what we know about even and odd functions, we notice that if we change one of the signs we can make something cancel out:

sin(θ−φ)=sin(θ)cos(−φ)+sin(−φ)cos(θ)sin⁡(θ−φ)=sin⁡(θ)cos⁡(−φ)+sin⁡(−φ)cos⁡(θ)
                        =sin(θ)cos(φ)−sin(φ)cos(θ)=sin⁡(θ)cos⁡(φ)−sin⁡(φ)cos⁡(θ)

sin(θ+φ)−sin(θ−φ)=2sin(θ)cos(φ)sin⁡(θ+φ)−sin⁡(θ−φ)=2sin⁡(θ)cos⁡(φ)

But now we have a formula for the difference of any two sines.  (Why?)
Written 1 Apr 2013 · View Upvotes
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  * Why is cosαsinβ=12[sin(α+β)−sin(α−β)]cos⁡αsin⁡β=12[sin⁡(α+β)−sin⁡(α−β)]?
  * Why is sinα+sinβ=2sin(α+β2)cos(α−β2)sin⁡α+sin⁡β=2sin⁡(α+β2)cos⁡(α−β2)?
  * Why is cosα−cosβ=−2sin(α+β2)sin(α−β2)cos⁡α−cos⁡β=−2sin⁡(α+β2)sin⁡(α−β2)?
  * Why is sinαsinβ=12[cos(α−β)−cos(α+β)]sin⁡αsin⁡β=12[cos⁡(α−β)−cos⁡(α+β)]?
  * Why is cosαcosβ=12[cos(α−β)+cos(α+β)]cos⁡αcos⁡β=12[cos⁡(α−β)+cos⁡(α+β)]?

William Mccoy, A former high school math teacher (Grades 9, 10, and 11). I have a Bachelor o...
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Manipulating the right side of the given identity:

  sin α ‒ sin β = 2{cos[(α + β)/2]sin[(α ‒ β)/2]}, we get:
 
1.)                  = 2{cos[(α/2) + (β/2)]sin[(α/2) ‒ (β/2)]}
 
We’ll now use The Sum of Two Angles formula for the cosine function and The Difference of Two Angles formula for the sine function:

cos[(α/2) + (β/2)] = cos (α/2) cos (β/2) ‒ sin (α/2) sin (β/2) 
sin[(α/2) ‒ (β/2)] = sin (α/2) cos (β/2) ‒ cos...

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