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Question: Solve \[\cos ix+i\sin ix\]= 1\. \[{{e}^{ix}}\] 2\. \[{{e}^{-ix}}\] 3\. \[{{e}^{x}}\] 4\. \[{...

Solve cosix+isinix\cos ix+i\sin ix=
1. eix{{e}^{ix}}
2. eix{{e}^{-ix}}
3. ex{{e}^{x}}
4. ex{{e}^{-x}}

Explanation

Solution

Hint : We can find the answer to this using the identities of trigonometric forms of complex numbers, this being a form of euler’s identity makes it easier for us to solve it. We can find the answer to this by using Taylor’s series and Maclaurin's expansions of trigonometric functions.

Complete step-by-step answer :
First to find the answer of this expression we need to write all of these identities in the taylor’s series defined for sine and cosine functions
Now as we know that
sinx=xx33!+x55!....\sin x=x-\dfrac{{{x}^{3}}}{3!}+\dfrac{{{x}^{5}}}{5!}-....
Now since we need the series for ix. Putting ix in place of x we get
sinix=ix(ix)33!+(ix)55!....\sin ix=ix-\dfrac{{{(ix)}^{3}}}{3!}+\dfrac{{{(ix)}^{5}}}{5!}-....
We can write this as
sinix=i(x+x33!+x55!+....)\sin ix=i\left( x+\dfrac{{{x}^{3}}}{3!}+\dfrac{{{x}^{5}}}{5!}+.... \right)
Now for the cosine function when we write the taylor’s series we get
cosx=1x22!+x44!....\cos x=1-\dfrac{{{x}^{2}}}{2!}+\dfrac{{{x}^{4}}}{4!}-....
Now we need to write this series for ix. Therefore substituting the value we get
cosix=1(ix)22!+(ix)44!....\cos ix=1-\dfrac{{{(ix)}^{2}}}{2!}+\dfrac{{{(ix)}^{4}}}{4!}-....
Solving this we get
cosix=1+x22!+x44!+....\cos ix=1+\dfrac{{{x}^{2}}}{2!}+\dfrac{{{x}^{4}}}{4!}+....
Now adding both to get the expression we need which is cosix+isinix\cos ix+i\sin ix we take and write it as
\cos ix+i\sin ix=\left( 1+\dfrac{{{x}^{2}}}{2!}+\dfrac{{{x}^{4}}}{4!}+.... \right)+i\left\\{ i\left( x+\dfrac{{{x}^{3}}}{3!}+\dfrac{{{x}^{5}}}{5!}+.... \right) \right\\}
When we open the parenthesis
cosix+isinix=(1+x22!+x44!+....)(x+x33!+x55!+....)\cos ix+i\sin ix=\left( 1+\dfrac{{{x}^{2}}}{2!}+\dfrac{{{x}^{4}}}{4!}+.... \right)-\left( x+\dfrac{{{x}^{3}}}{3!}+\dfrac{{{x}^{5}}}{5!}+.... \right)
Subtracting both the brackets
cosix+isinix=1x+x22!x33!+x44!x55!+....\cos ix+i\sin ix=1-x+\dfrac{{{x}^{2}}}{2!}-\dfrac{{{x}^{3}}}{3!}+\dfrac{{{x}^{4}}}{4!}-\dfrac{{{x}^{5}}}{5!}+....
Now since we want to write this in the form of an expansion we can write the expansion of ex{{e}^{x}}. Now we know that the expansion of ex{{e}^{x}} is
ex=1+x+x22!+x33!+x44!+x55!+....{{e}^{x}}=1+x+\dfrac{{{x}^{2}}}{2!}+\dfrac{{{x}^{3}}}{3!}+\dfrac{{{x}^{4}}}{4!}+\dfrac{{{x}^{5}}}{5!}+....
Now if we substitute the negative of x instead of x we get
ex=1+(x)+(x)22!+(x)33!+(x)44!+(x)55!+....{{e}^{-x}}=1+(-x)+\dfrac{{{(-x)}^{2}}}{2!}+\dfrac{{{(-x)}^{3}}}{3!}+\dfrac{{{(-x)}^{4}}}{4!}+\dfrac{{{(-x)}^{5}}}{5!}+....
Now when we open the brackets we get
ex=1x+x22!x33!+x44!x55!+....{{e}^{-x}}=1-x+\dfrac{{{x}^{2}}}{2!}-\dfrac{{{x}^{3}}}{3!}+\dfrac{{{x}^{4}}}{4!}-\dfrac{{{x}^{5}}}{5!}+....
This as we know is equal to the solution we need, which is cosix+isinix\cos ix+i\sin ix.
Hence we can say the answer of the expression cosix+isinix\cos ix+i\sin ix is ex{{e}^{-x}} which is option d.
So, the correct answer is “Option 2”.

Note: Taylor series can be explained as being the representation of any function in the form of an infinite series of terms calculated from the values of the function’s derivatives at a single point. We can remember the result as a theorem which can be used further.