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Question: Multiply \({{e}^{\sqrt{-1}}}+{{e}^{-\sqrt{-1}}}\) by \({{e}^{\sqrt{-1}}}-{{e}^{-\sqrt{-1}}}\)....

Multiply e−1+e−−1{{e}^{\sqrt{-1}}}+{{e}^{-\sqrt{-1}}} by e−1−e−−1{{e}^{\sqrt{-1}}}-{{e}^{-\sqrt{-1}}}.

Explanation

Solution

Hint:Here, (e−1+e−−1)(e−1−e−−1)\left( {{e}^{\sqrt{-1}}}+{{e}^{-\sqrt{-1}}} \right)\left( {{e}^{\sqrt{-1}}}-{{e}^{-\sqrt{-1}}} \right) is of the form (a+b)(a−b)(a+b)(a-b) which is a2−b2{{a}^{2}}-{{b}^{2}} and we have to substitute for −1=i\sqrt{-1}=i, and then apply the formula eiθ=cos⁡θ+isin⁡θ{{e}^{i\theta }}=\cos \theta +i\sin \theta and hence, do the simplification.

Complete Step-by-step answer:
Here, we have to calculate (e−1+e−−1)(e−1−e−−1)\left( {{e}^{\sqrt{-1}}}+{{e}^{-\sqrt{-1}}} \right)\left( {{e}^{\sqrt{-1}}}-{{e}^{-\sqrt{-1}}} \right).
That is, it is of the form (a+b)(a−b)(a+b)(a-b). We know the formula that:
(a+b)(a−b)=a2−b2(a+b)(a-b)={{a}^{2}}-{{b}^{2}}
Here, we have a=e−1a={{e}^{\sqrt{-1}}} and b=e−−1b={{e}^{-\sqrt{-1}}}.
Therefore, by applying the formula we get the equation:
(e−1+e−−1)(e−1−e−−1)=(e−1)2−(e−−1)2\left( {{e}^{\sqrt{-1}}}+{{e}^{-\sqrt{-1}}} \right)\left( {{e}^{\sqrt{-1}}}-{{e}^{-\sqrt{-1}}} \right)={{\left( {{e}^{\sqrt{-1}}} \right)}^{2}}-{{\left( {{e}^{-\sqrt{-1}}} \right)}^{2}} …. (1)
We know that −1\sqrt{-1} is a complex number and it is considered as the imaginary value ii. Hence, we can write:
−1=i\sqrt{-1}=i
Therefore, our equation (1) becomes:
(e−1+e−−1)(e−1−e−−1)=(ei)2−(e−i)2\left( {{e}^{\sqrt{-1}}}+{{e}^{-\sqrt{-1}}} \right)\left( {{e}^{\sqrt{-1}}}-{{e}^{-\sqrt{-1}}} \right)={{\left( {{e}^{i}} \right)}^{2}}-{{\left( {{e}^{-i}} \right)}^{2}} …. (2)
We also know the property that (am)n=amn{{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}.
Therefore, we can say that (ei)2=e2i{{\left( {{e}^{i}} \right)}^{2}}={{e}^{2i}} and (e−i)2=e−2i{{\left( {{e}^{-i}} \right)}^{2}}={{e}^{-2i}}.
Hence, our equation (2) becomes:
(e−1+e−−1)(e−1−e−−1)=e2i−e−2i\left( {{e}^{\sqrt{-1}}}+{{e}^{-\sqrt{-1}}} \right)\left( {{e}^{\sqrt{-1}}}-{{e}^{-\sqrt{-1}}} \right)={{e}^{2i}}-{{e}^{-2i}} …. (3)
The RHS in the above equation is of the form eiθ{{e}^{i\theta }}.
We know the formula that:
eiθ=cos⁡θ+isin⁡θ{{e}^{i\theta }}=\cos \theta +i\sin \theta
We have e2i=ei2{{e}^{2i}}={{e}^{i2}}
By applying this formula we will get:
e2i=cos⁡2+isin⁡2{{e}^{2i}}=\cos 2+i\sin 2 …… (4)
Similarly, we can say that,
e−2i=ei(−2) e−2i=cos⁡(−2)+isin⁡(−2) \begin{aligned} & {{e}^{-2i}}={{e}^{i(-2)}} \\\ & {{e}^{-2i}}=\cos (-2)+i\sin (-2) \\\ \end{aligned}
We know that cos⁡(−x)=cos⁡x\cos (-x)=\cos x and sin⁡(−x)=−sin⁡x\sin (-x)=-\sin x. Hence, we will get:
cos⁡(−2)=cos⁡2\cos (-2)=\cos 2 and sin⁡(−2)=−sin⁡2\sin (-2)=-\sin 2
Therefore, our equation becomes:
e−2i=cos⁡2−isin⁡2{{e}^{-2i}}=\cos 2-i\sin 2 ….. (5)
Hence, by substituting equation (4) and equation (5) in equation (3) we obtain the equation:
(e−1+e−−1)(e−1−e−−1)=cos⁡2+isin⁡2−(cos⁡2−isin⁡2)\left( {{e}^{\sqrt{-1}}}+{{e}^{-\sqrt{-1}}} \right)\left( {{e}^{\sqrt{-1}}}-{{e}^{-\sqrt{-1}}} \right)=\cos 2+i\sin 2-(\cos 2-i\sin 2)
Now, take cos⁡2−isin⁡2\cos 2-i\sin 2outside the bracket we get:
(e−1+e−−1)(e−1−e−−1)=cos⁡2+isin⁡2−cos⁡2+isin⁡2\left( {{e}^{\sqrt{-1}}}+{{e}^{-\sqrt{-1}}} \right)\left( {{e}^{\sqrt{-1}}}-{{e}^{-\sqrt{-1}}} \right)=\cos 2+i\sin 2-\cos 2+i\sin 2
In the next step, we have to cancel cos⁡2\cos 2 and isin⁡2+isin⁡2=2isin⁡2i\sin 2+i\sin 2=2i\sin 2, so we obtain the equation: (e−1+e−−1)(e−1−e−−1)=2isin⁡2\left( {{e}^{\sqrt{-1}}}+{{e}^{-\sqrt{-1}}} \right)\left( {{e}^{\sqrt{-1}}}-{{e}^{-\sqrt{-1}}} \right)=2i\sin 2
Therefore, by multiplying e−1+e−−1{{e}^{\sqrt{-1}}}+{{e}^{-\sqrt{-1}}} by e−1−e−−1{{e}^{\sqrt{-1}}}-{{e}^{-\sqrt{-1}}} we obtain the solution, 2isin⁡22i\sin 2, where ii is the imaginary number whose value is taken as, i=−1i=\sqrt{-1}.

Note: Here, after getting (e−1+e−−1)(e−1−e−−1)=e2i−e−2i\left( {{e}^{\sqrt{-1}}}+{{e}^{-\sqrt{-1}}} \right)\left( {{e}^{\sqrt{-1}}}-{{e}^{-\sqrt{-1}}} \right)={{e}^{2i}}-{{e}^{-2i}}, you should not stop. Next, you change this equation into the form eiθ=cos⁡θ+isin⁡θ{{e}^{i\theta }}=\cos \theta +i\sin \theta and reduce it into a much simpler form. You also should have an idea regarding the properties of exponential functions.