### The Log-Antilog Procedure: Part II

Not too long later, I learned in additional mathematics, certain trigonometric identities of the form cos(A ± B) = –(cos A cos B ± sin A sin B), from which we can form another identity: cos A cos B = ½ [cos(A + B) + cos(A – B)].

This identity is the basis of an obsolete technique known as prosthaphaeretic multiplication, which is very similar but older than the log-antilog procedure taught by my teacher. To illustrate prosthaphaeretic multiplication on 10.8 x 87.85, we proceeds as follows: First write 10.8 x 87.85 as 0.108 x 0.8785 x 10000, then read the arccosine values for 0.108 and 0.8785 from table, they are 1.463 and 0.498 in radians. With the identity, multiplication of two cosines is converted to addition/subtraction of their respective arccosines, that is,

10.8 x 87.85 = cos 1.463 x cos 0.498 x 10000 = ½[cos(1.463 + 0.498) + cos(1.463 - 0.498)] x 10000

The result is 945, which is very close to the exact solution: 948.78.

At the time, I thought, the only sensible reason for retain…

This identity is the basis of an obsolete technique known as prosthaphaeretic multiplication, which is very similar but older than the log-antilog procedure taught by my teacher. To illustrate prosthaphaeretic multiplication on 10.8 x 87.85, we proceeds as follows: First write 10.8 x 87.85 as 0.108 x 0.8785 x 10000, then read the arccosine values for 0.108 and 0.8785 from table, they are 1.463 and 0.498 in radians. With the identity, multiplication of two cosines is converted to addition/subtraction of their respective arccosines, that is,

10.8 x 87.85 = cos 1.463 x cos 0.498 x 10000 = ½[cos(1.463 + 0.498) + cos(1.463 - 0.498)] x 10000

The result is 945, which is very close to the exact solution: 948.78.

At the time, I thought, the only sensible reason for retain…