What is the trigonometric form of #5e^(pi i) #?
1 Answer
Jan 23, 2016
# 5 [cos(pi) + isin(pi) ] #
Explanation:
Euler's theorem states that :
# e^(itheta) = costheta + isintheta #
# rArr5e^(pi i) = 5[ cos(pi) + isin(pi) ] # Note : since
# cos(pi) = - 1 and sin(pi) = 0# the expression may be simplified to - 5