What is e^{ix}+e^{i(x+pi/8)} ?

The solution to e^{ix}+e^{i(x+pi/8)} is (cos(x)+cos((pi+8x)/8))+i(sin((pi+8x)/8)+sin(x))

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simplify e^{ix}+e^{i(x+pi/8)}

simplify

e^{i x} + e^{i (x + \frac{π}{8})}

(cos (x) + cos (\frac{π + 8 x}{8})) + i (sin (\frac{π + 8 x}{8}) + sin (x))

Solution steps

e^{i x} + e^{i (x + \frac{π}{8})}

Apply imaginary number rule:

e^{ia} = cos (a) + i sin (a)

= cos (x) + i sin (x) + e^{i (x + \frac{π}{8})}

Apply imaginary number rule:

e^{ia} = cos (a) + i sin (a)

= cos (x) + i sin (x) + cos (\frac{8 x + π}{8}) + i sin (\frac{8 x + π}{8})

Rewrite

cos (x) + sin (x) i + cos (\frac{8 x + π}{8}) + sin (\frac{8 x + π}{8}) i

in standard complex form:

(cos (x) + cos (\frac{8 x + π}{8})) + (sin (x) + sin (\frac{8 x + π}{8})) i

= (cos (x) + cos (\frac{8 x + π}{8})) + (sin (x) + sin (\frac{8 x + π}{8})) i

\frac{3 ( 1 + 5 e ^{6 x} )}{1 - 5 e ^{6 x}}

\frac{d}{d s} ((K (s)) (s + 3))

3 (5 c - 4 = 8 c + 2)

\frac{( 6 - y ) ^{3}}{3 y}

\frac{2 ( 2 x - 1 )}{3 ( x + 1 )}

What is e^{ix}+e^{i(x+pi/8)} ?
The solution to e^{ix}+e^{i(x+pi/8)} is (cos(x)+cos((pi+8x)/8))+i(sin((pi+8x)/8)+sin(x))