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How do you find an expression sin(x) in terms of eix and eix ?

Answer
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Hint:
In this question, we were asked to find the expression of sin x in terms of eix and eix . So, here we will use the McLaurin formula from the series of the exponential function. We will also use the basic properties of trigonometry. So let us see how we can solve this problem.

Complete step by step solution:
For solving the above question, we will use the McLaurin formula which is ex=n=0xnn!
So, ex=n=0(ix)nn!=n=0inxnn!
Let us suppose that n = 2k in the first case and n = 2k + 1 in the second case, so that we can separate the odd and even terms for n
Now see,
 i2k=(i2)k=(1)k
So, eix=n=0(1)kx2n(2k)!+in=0(1)kx2k+1(2k+1)!
We know that the McLaurin expansions of cos x and sin x, and thus we get
 eix=cosx+isinx
Which is Euler’s formula
Now, let us suppose that cos x is an even function and sin x is an odd function then we will get,
 eix=cos(x)+isin(x)=cosxisinx
 eixeix=2isinx
Finally, we will get
 sinx=eixeix2i

Therefore, on solving sin x in terms of eix and eix we get sinx=eixeix2i

Note:
There is an alternative way, to solve this problem, let us see that as well.
We know that eix=cosx+isinx (from Euler’s formula)
Also, eix=cos(x)+isin(x)
Since, cos(-x) = cos x and sin(-x) = -sin x
Now, we have
eix=cosx+isinx
eix=cosxisinx
On adding both of them we get,
eix+eix=2cosx
Now, cosx=eixeix2
On subtracting both of them we get,
eixeix=2isinx
Then, sinx=eixeix2i
Therefore, even with this alternative way, we get the same solution that is sinx=eixeix2i