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Question: How do you find an expression sin(x) in terms of \({{e}^{ix}}\) and \({{e}^{ix}}\) ?...

How do you find an expression sin(x) in terms of eix{{e}^{ix}} and eix{{e}^{ix}} ?

Explanation

Solution

In this question, we were asked to find the expression of sin x in terms of eix{{e}^{ix}} and eix{{e}^{ix}} . 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=0∞xnn!{{e}^{x}}=\sum\limits_{n=0}^{\infty }{\dfrac{x{}^{n}}{n!}}
So, ex=∑n=0∞(ix)nn!=∑n=0∞inxnn!{{e}^{x}}=\sum\limits_{n=0}^{\infty }{\dfrac{(ix){}^{n}}{n!}}=\sum\limits_{n=0}^{\infty }{i{}^{n}\dfrac{x{}^{n}}{n!}}
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{{i}^{2k}}={{({{i}^{2}})}^{k}}={{(-1)}^{k}}
So, eix=∑n=0∞(−1)kx2n(2k)!+i∑n=0∞(−1)kx2k+1(2k+1)!{{e}^{ix}}=\sum\limits_{n=0}^{\infty }{{{(-1)}^{k}}\dfrac{x{}^{2n}}{(2k)!}+}i\sum\limits_{n=0}^{\infty }{{{(-1)}^{k}}\dfrac{x{}^{2k+1}}{(2k+1)!}}
We know that the McLaurin expansions of cos x and sin x, and thus we get
eix=cos⁡x+isin⁡x{{e}^{ix}}=\cos x+i\sin x
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,
e−ix=cos⁡(−x)+isin⁡(−x)=cos⁡x−isin⁡x{{e}^{-ix}}=\cos (-x)+i\sin (-x)=\cos x-i\sin x
⇒eix−e−ix=2isin⁡x\Rightarrow {{e}^{ix}}-{{e}^{-ix}}=2i\sin x
Finally, we will get
⇒sin⁡x=eix−e−ix2i\Rightarrow \sin x=\dfrac{{{e}^{ix}}-{{e}^{-ix}}}{2i}

Therefore, on solving sin x in terms of eix{{e}^{ix}} and eix{{e}^{ix}} we get sin⁡x=eix−e−ix2i\sin x=\dfrac{{{e}^{ix}}-{{e}^{-ix}}}{2i}

Note:
There is an alternative way, to solve this problem, let us see that as well.
We know that eix=cos⁡x+isin⁡x{{e}^{ix}}=\cos x+i\sin x (from Euler’s formula)
Also, e−ix=cos⁡(−x)+isin⁡(−x){{e}^{-ix}}=\cos (-x)+i\sin (-x)
Since, cos(-x) = cos x and sin(-x) = -sin x
Now, we have
eix=cos⁡x+isin⁡x{{e}^{ix}}=\cos x+i\sin x
e−ix=cos⁡x−isin⁡x{{e}^{-ix}}=\cos x-i\sin x
On adding both of them we get,
⇒eix+e−ix=2cos⁡x\Rightarrow {{e}^{ix}}+{{e}^{-ix}}=2\cos x
Now, ⇒cos⁡x=eix−e−ix2\Rightarrow \cos x=\dfrac{{{e}^{ix}}-{{e}^{-ix}}}{2}
On subtracting both of them we get,
eix−e−ix=2isin⁡x{{e}^{ix}}-{{e}^{-ix}}=2i\sin x
Then, sin⁡x=eix−e−ix2i\sin x=\dfrac{{{e}^{ix}}-{{e}^{-ix}}}{2i}
Therefore, even with this alternative way, we get the same solution that is sin⁡x=eix−e−ix2i\sin x=\dfrac{{{e}^{ix}}-{{e}^{-ix}}}{2i}