(M-11)  Deriving   sin(a+b),   cos(a+b)

Given the functions (sina, cosa, sinb and cos b), we seek formulas that express sin(a+b) and cos(a+b). The first of these formulas is used in deriving the L4 and L5 Lagrangian points, here.

Please verify every calculation step before proceeding!

As shown in the drawing, to derive the formula we combine two right-angled triangles

    ABC which has an angle a
    ABD which " ""b
The long side ("hypotenuse') of ACD is AD=R. Therefore

    DC = R sin b
    AC = R cos b
Similarly

    BC = AC sin a = R cos b sin a
    AB = AC cos a = R cos b cos a
The triangle ADF is right-angled and has the angle (a+b). Therefore

    R sin (a+b) = DF
    R cos (a+b) = AF

Start by deriving the sine:

    R sin (a+b) = DF  =  EF + DE  =  BC + DE
  Note in the drawing the two head-to-head angles marked with double lines: like all such angles, they must be equal. Each of them is one of the two sharp ("acute") angles in its own right-angled triangle. Since the sharp angles in such a triangle add up to 90 degrees, the other two sharp angles must be equal. This justifies marking the angle near D as a, as drawn in the figure.

In the right-angled triangle CED

    DE = DC cos a = R sin b cos a
    EC = DC sin a = R sin b sin a
Earlier it was already shown that
    BC = R cos b sin a
    AB = R cos b cos a
Therefore

 R sin (a+b)  =  BC+DE  =  R cos b sin a + R sin b cos a

Cancelling R and rearranging a to precede b

   sin (a+b)  =  cos b sin a + sin b cos a


Similarly, for the cosine

R cos (a+b) = AF = AB - FB = AB - EC =

        = R cos b cos a - R sin b sin a

Cancelling R and rearranging

   cos (a+b)  =  cos a cos b - R sin a sin b


The L4 and L5 Lagrangian Points



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Author and curator: David P. Stern
Last updated 3 April 1999


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