Question 799298
<pre>
[sin(a)+cos(a)] + i[sin(a)-cos(a)]

We use the facts that

1. cos(X)cos(Y)+cos(X)sin(Y) = cos(X-Y)
2. sin(X)cos(Y)-cos(X)sin(Y) = sin(X-Y)

and

3. sin({{{pi/4}}}) = cos({{{pi/4}}}) = {{{1/sqrt(2)}}}

Let's take the real part first:

sin(a)+cos(a), write it over 1, {{{(sin(a)+cos(a))/1}}}   

Multiply it by {{{cos(pi/4)/cos(pi/4)}}} which just equals 1
and therefore will not change the value:

{{{cos(pi/4)/cos(pi/4)}}}{{{""*""}}}{{{(sin(a)+cos(a))/1}}}

{{{( cos(pi/4)(sin(a)+cos(a)))/cos(pi/4)}}} 

Distribute on top 

{{{( cos(pi/4)sin(a)+cos(pi/4)cos(a) )/cos(pi/4)}}}

By 3 above, replace the first cos({{{pi/4}}}) by sin({{{pi/4}}})

{{{( sin(pi/4)sin(a)+cos(pi/4)cos(a) )/cos(pi/4)}}}

Rearrange to look like cos(X)cos(Y)+cos(X)sin(Y) = cos(X-Y)

{{{( cos(a)cos(pi/4)+sin(a)sin(pi/4) )/cos(pi/4)}}} 

So the numerator becomes cos(a-{{{pi/4}}})

{{{cos(a-pi/4)/cos(pi/4)}}} 

Since the denominator {{{cos(pi/4)}}} = {{{1/sqrt(2)}}}, we substitute 
and get:

{{{cos(a-pi/4)/(1/sqrt(2))}}} = {{{cos(a-pi/4)}}}{{{"÷"}}}{{{1/sqrt(2)}}} = {{{cos(a-pi/4)}}}{{{""*""}}}{{{sqrt(2)/1}}



So the real part becaomes

{{{sqrt(2)}}}{{{cos(a-pi/4)}}} 

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Now we take the imaginary part, the coefficient of i:

sin(a)-cos(a), write it over 1, {{{(sin(a)-cos(a))/1}}}   

Multiply it by {{{sin(pi/4)/sin(pi/4)}}} which just equals 1
and therefore will not change the value:

{{{sin(pi/4)/sin(pi/4)}}}{{{""*""}}}{{{(sin(a)-cos(a))/1}}}

{{{( sin(pi/4)(sin(a)-cos(a)))/sin(pi/4)}}} 

Distribute on top 

{{{( sin(pi/4)sin(a)-sin(pi/4)cos(a) )/sin(pi/4)}}}

By 3 above, replace the first sin({{{pi/4}}}) by cos({{{pi/4}}})

{{{( cos(pi/4)sin(a)-sin(pi/4)cos(a) )/sin(pi/4)}}}

Rearrange to look like sin(X)cos(Y)-cos(X)sin(Y) = sin(X-Y)

{{{( sin(a)cos(pi/4)-cos(a)sin(pi/4) )/sin(pi/4)}}} 

So the numerator becomes sin(a-{{{pi/4}}})

{{{sin(a-pi/4)/sin(pi/4)}}} 

Since the denominator {{{sin(pi/4)}}} = {{{1/sqrt(2)}}}, we substitute 
and get:

{{{sin(a-pi/4)/(1/sqrt(2))}}} = {{{sin(a-pi/4)}}}{{{"÷"}}}{{{1/sqrt(2)}}} = {{{sin(a-pi/4)}}}{{{""*""}}}{{{sqrt(2)/1}}

So the imaginary part becaomes

{{{sqrt(2)}}}{{{sin(a-pi/4)}}} 

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So the original problem 

[sin(a)+cos(a)] + i[sin(a)-cos(a)]

becomes:

{{{sqrt(2)}}}{{{cos(a-pi/4)}}} + i·{{{sqrt(2)}}}{{{sin(a-pi/4)}}} 

{{{sqrt(2)}}}[{{{cos(a-pi/4)}}} + i·{{{sin(a-pi/4)}}}] 

That's the polar representation

which is often written as 

{{{sqrt(2)}}}{{{cis(a-pi/4)}}},

and electrical engineers

write it as {{{sqrt(2)}}}&#8736;{{{(a-pi/4)}}}.  

 Edwin</pre>