What trigonometric identity was used to get from #sin((kpix)/L)[A_kcos((kpict)/L) + B_ksin((kpict)/L)]# to #E_ksin((kpix)/L)cos[(kpic)/L(t-t_k)]#? What is the relationship between the pair of constants #E_k# and #t_k#, and the pair #A_k# and #B_k#?

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This is something from Zachmanoglou and Thou, and they don't say which identity is used or show how it was done.

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Answer 1 · 2017-03-31T02:14:35

Any expression of the form:

# AsinX + BcosX #

can be put in the form:

#Rcos(X-alpha)#

If we use the #cos# sum formula then we have:

# Rcos(X-alpha) -= R{ cosXcosalpha + sin X sin alpha }#

So we require:

# AsinX + BcosX = R{ cosXcosalpha + sin X sin alpha }#
# :. AsinX + BcosX = R sin alpha sin X + RcosalphacosX #

Equating coefficient gives us two equatons;

# R sin alpha = A # .... [1]
# R cosalpha = B # .... [2]

#Eq [1]^2 + Eq[2]^2:#

# R^2 sin^2alpha+R^2cos^2alpha=A^2+B^2 #
# :. R^2 (sin^2alpha+cos^2alpha)=A^2+B^2 #
# :. R^2 =A^2+B^2 #

#Eq [1] -: Eq[2]:#

# (R sin alpha)/(R cosalpha) = A/B #
# :. tan alpha = A/B #
# :. alpha = tan^(-1)(A/B) #

Here;

# R=E_k #
# A=B_k #
# B=A_k #
# alpha = (kpic)/L t_k#