# 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.

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

##### 1 Answer

Any expression of the form:

# AsinX + BcosX #

can be put in the form:

#Rcos(X-alpha)#

If we use the

# 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]

# 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 #

# (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#