__cos b = 1/2 [cos (a+b) + cos (a-b)]
a. sin(a)
b. sin(b)
c. cos(a)
d. cos(b...
Mathematics, 30.01.2020 10:56, shartman22
__cos b = 1/2 [cos (a+b) + cos (a-b)]
a. sin(a)
b. sin(b)
c. cos(a)
d. cos(b)
Answers: 2
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