Exponential forms of cos and sin

Dough
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hi, my question is from Modern Engineering Mathematics by Glyn James

pg 177 # 17a

Using the exponential forms of cos(theta) and sin(theta) given in (3.11a, b), prove the following trigonometric identities:
a) sin(x + y) = sin(x)cos(y) + cos(x)sin(y)

and 3.11a is:
cos(x) = 0.5*[ e^(jx) + e^(-jx) ] where x= theta
and 3.11b is:
sin(x) = 0.5j*[ e^(jx) - e^(-jx) ] where x= theta

i've gotten to the point where i have
[ e^j(x+y) + e^j(x-y) -e^j(y-x) -e^-j(x + y) ] / 2j
 
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3.11b should be

sin(x) = 0.5*[ e^(jx) - e^(-jx) ]/j

or

sin(x) = -0.5j*[ e^(jx) - e^(-jx) ]

What you want to do here is start with the expression 3.11b of sin(x+y), then use the property of the exponential that exp(x+y)=exp(x)exp(y) and then transform the exponentials back into sin and cos form. Then the real part of that is sin(x+y) (and the imaginary part equals 0).
 
whoops i meant for the j to be on the bottom for 3.11b

its like
[1/(2j)][ e^(jx) - e^(-jx) ]
 
weee i got it thanks for the help!
 
Oh, those engineers and their jmagjnary numbers!
 
hello my physics chums,
hows your equations looking? I want somebody to love me.

kevmyster
 
HallsofIvy said:
Oh, those engineers and their jmagjnary numbers!

Yes that's a pain in the ..., always confusion between i and j.
 
HallsofIvy said:
Oh, those engineers and their jmagjnary numbers!

In electrical engineering, i means electric current so imaginary numbers are written as j. I remember j was first introduced in the "Elementary linear circuit analysis" class and then all engineering classes use j instead of i. Yes, when reading mathematics or physics paper, I need to switch to i mode. And, mathematicians and physicists write Fourier transform in a confusing form
 
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