There are many arguments as for why the equality holds, but one good way of understanding for a calculus student comes from looking at the Taylor series' for the exponential, sine, and cosine functions.
#e^x = sum_(n=0)^oo x^n/(n!)#
#sin(x) = sum_(n=0)^oo (-1)^nx^(2n+1)/((2n+1)!)#
#cos(x) = sum_(n=0)^oo (-1)^nx^(2n)/((2n)!)#
Starting from the left hand side:
#isin(x)+cos(x) = isum_(n=0)^oo (-1)^nx^(2n+1)/((2n+1)!) + sum_(n=0)^oo (-1)^nx^(2n)/((2n)!)#
#= (ix/(1!)-ix^3/(3!) + ix^5/(5!) - ... ) + (1 - x^2/(2!) + x^4/(4!) - ...)#
#= 1 + ix/(1!) -x^2/(2!) -ix^3/(3!) +x^4/(4!) + ix^5/(5!) -...#
#= i^0 1 + i^1x/(1!) +i^2x^2/(2!) +i^3x^3/(3!) +i^4x^4/(4!) + i^5x^5/(5!) -...#
#= sum_(n=0)^ooi^nx^n/(n!)#
#=sum_(n=0)^oo (ix)^n/(n!)#
#=e^x#
Multiplying each side by #a# gives the desired result.
Another accessible argument is based around integration
Let #u = isin(x)+cos(x)#
#=> du = (-sin(x)+icos(x))dx#
#= i(isin(x)+cos(x))dx#
#= iudx#
#=> 1/udu = idx#
#=> int1/udu = intidx#
#=> lnu = ix#
#=> u = e^ix#
#=> isin(x) + cos(x) = e^ix#
These arguments are meant to show that if we extend our typical tools into the complex plane and let them work the same way, then the given identity is the result.
