Answer

Verified

449.4k+ views

Hint: Use Euler’s identity ${{e}^{i\theta }}=\cos \theta +i\sin \theta $and ${{i}^{2}}=-1$. Try converting the numerator and denominator in $\cos \theta +i\sin \theta $ form. Use the trigonometric identities $1+\cos \theta =2{{\cos }^{2}}\dfrac{\theta }{2}$ and $\sin \theta =2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}$.

Complete step-by-step answer:

Let \[L=\dfrac{1+\sin \dfrac{\pi }{8}+i\cos \dfrac{\pi }{8}}{1+\sin \dfrac{\pi }{8}-i\cos \dfrac{\pi }{8}}\]

Using $\sin x=\cos \left( \dfrac{\pi }{2}-x \right)$ and $\cos x=\sin \left( \dfrac{\pi }{2}-x \right)$ we get

\[L=\dfrac{1+\cos \dfrac{3\pi }{8}+i\sin \dfrac{3\pi }{8}}{1+\cos \dfrac{3\pi }{8}-i\sin \dfrac{3\pi }{8}}\]

Using $1+\cos \theta =2{{\cos }^{2}}\dfrac{\theta }{2}$ and $\sin \theta =2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}$ we get

$L=\dfrac{2{{\cos }^{2}}\dfrac{3\pi }{16}+2i\sin \dfrac{3\pi }{16}\cos \dfrac{3\pi }{16}}{2{{\cos }^{2}}\dfrac{3\pi }{16}-2i\sin \dfrac{3\pi }{16}\cos \dfrac{3\pi }{16}}$

Taking $2\cos \dfrac{3\pi }{16}$ common from numerator and denominator, we get

$L=\dfrac{2\cos \dfrac{3\pi }{16}\left( \cos \dfrac{3\pi }{16}+i\sin \dfrac{3\pi }{16} \right)}{2\cos \dfrac{3\pi }{16}\left( \cos \dfrac{3\pi }{16}-i\sin \dfrac{3\pi }{16} \right)}$

Simplifying we get

$\begin{align}

& L=\dfrac{{{e}^{i\dfrac{3\pi }{16}}}}{{{e}^{-i\dfrac{3\pi }{16}}}} \\

& ={{e}^{i\dfrac{3\pi }{16}+i\dfrac{3\pi }{16}}} \\

& ={{e}^{i\dfrac{6\pi }{16}}} \\

& ={{e}^{\dfrac{3\pi }{8}}} \\

\end{align}$

Hence ${{\left( \dfrac{1+\sin \dfrac{\pi }{8}+i\cos \dfrac{\pi }{8}}{1+\sin \dfrac{\pi }{8}-i\cos \dfrac{\pi }{8}} \right)}^{\dfrac{8}{3}}}={{L}^{\dfrac{8}{3}}}={{e}^{i\dfrac{3\pi }{8}\times \dfrac{8}{3}}}={{e}^{i\pi }}=\cos \pi +i\sin \pi =-1$

Hence, we have ${{\left( \dfrac{1+\sin \dfrac{\pi }{8}+i\cos \dfrac{\pi }{8}}{1+\sin \dfrac{\pi }{8}-i\cos \dfrac{\pi }{8}} \right)}^{\dfrac{8}{3}}}=-1$.

Note:

[1] Euler’s identity ${{e}^{i\theta }}=\cos \theta +i\sin \theta $

[2] Demovire’s identity ${{\left( \cos x+i\sin x \right)}^{n}}=\cos nx+i\sin nx$

[3] We can use identity $\dfrac{1+\cos x+i\sin x}{1+\cos x-i\sin x}=\cos x+i\sin x$ directly to solve the question. The derivation of the above formula will follow the same steps as done in the above solution.

[4] Complex numbers help in finding the values of $\cos nx$ and $\sin nx$ conveniently and easily.

[5] Every complex number can be written in the form of $r{{e}^{ix}}$ where r is called the modulus of the complex number and x is called the argument of the complex number.

[6] Complex numbers play a role of backbone in calculus and are used in many places.

[7] $\left| z \right|=r$ is the equation of circle centred at (0,0) and radius r in the argand plane.

[8] Complex numbers are used in Calculus e.g. Cauchy residue theorem (Theorem for complex integrals) is used to evaluate some difficult real integrals.

[9] Multiplication by $i$ is equivalent to a rotation of $90{}^\circ $ in the argand plane.

[10] A complex number can be thought of as a vector in the Argand plane. This is why complex numbers follow a similar algebra as vector algebra.

[11] Complex numbers are used in AC circuit analysis etc in physics.

Complete step-by-step answer:

Let \[L=\dfrac{1+\sin \dfrac{\pi }{8}+i\cos \dfrac{\pi }{8}}{1+\sin \dfrac{\pi }{8}-i\cos \dfrac{\pi }{8}}\]

Using $\sin x=\cos \left( \dfrac{\pi }{2}-x \right)$ and $\cos x=\sin \left( \dfrac{\pi }{2}-x \right)$ we get

\[L=\dfrac{1+\cos \dfrac{3\pi }{8}+i\sin \dfrac{3\pi }{8}}{1+\cos \dfrac{3\pi }{8}-i\sin \dfrac{3\pi }{8}}\]

Using $1+\cos \theta =2{{\cos }^{2}}\dfrac{\theta }{2}$ and $\sin \theta =2\sin \dfrac{\theta }{2}\cos \dfrac{\theta }{2}$ we get

$L=\dfrac{2{{\cos }^{2}}\dfrac{3\pi }{16}+2i\sin \dfrac{3\pi }{16}\cos \dfrac{3\pi }{16}}{2{{\cos }^{2}}\dfrac{3\pi }{16}-2i\sin \dfrac{3\pi }{16}\cos \dfrac{3\pi }{16}}$

Taking $2\cos \dfrac{3\pi }{16}$ common from numerator and denominator, we get

$L=\dfrac{2\cos \dfrac{3\pi }{16}\left( \cos \dfrac{3\pi }{16}+i\sin \dfrac{3\pi }{16} \right)}{2\cos \dfrac{3\pi }{16}\left( \cos \dfrac{3\pi }{16}-i\sin \dfrac{3\pi }{16} \right)}$

Simplifying we get

$\begin{align}

& L=\dfrac{{{e}^{i\dfrac{3\pi }{16}}}}{{{e}^{-i\dfrac{3\pi }{16}}}} \\

& ={{e}^{i\dfrac{3\pi }{16}+i\dfrac{3\pi }{16}}} \\

& ={{e}^{i\dfrac{6\pi }{16}}} \\

& ={{e}^{\dfrac{3\pi }{8}}} \\

\end{align}$

Hence ${{\left( \dfrac{1+\sin \dfrac{\pi }{8}+i\cos \dfrac{\pi }{8}}{1+\sin \dfrac{\pi }{8}-i\cos \dfrac{\pi }{8}} \right)}^{\dfrac{8}{3}}}={{L}^{\dfrac{8}{3}}}={{e}^{i\dfrac{3\pi }{8}\times \dfrac{8}{3}}}={{e}^{i\pi }}=\cos \pi +i\sin \pi =-1$

Hence, we have ${{\left( \dfrac{1+\sin \dfrac{\pi }{8}+i\cos \dfrac{\pi }{8}}{1+\sin \dfrac{\pi }{8}-i\cos \dfrac{\pi }{8}} \right)}^{\dfrac{8}{3}}}=-1$.

Note:

[1] Euler’s identity ${{e}^{i\theta }}=\cos \theta +i\sin \theta $

[2] Demovire’s identity ${{\left( \cos x+i\sin x \right)}^{n}}=\cos nx+i\sin nx$

[3] We can use identity $\dfrac{1+\cos x+i\sin x}{1+\cos x-i\sin x}=\cos x+i\sin x$ directly to solve the question. The derivation of the above formula will follow the same steps as done in the above solution.

[4] Complex numbers help in finding the values of $\cos nx$ and $\sin nx$ conveniently and easily.

[5] Every complex number can be written in the form of $r{{e}^{ix}}$ where r is called the modulus of the complex number and x is called the argument of the complex number.

[6] Complex numbers play a role of backbone in calculus and are used in many places.

[7] $\left| z \right|=r$ is the equation of circle centred at (0,0) and radius r in the argand plane.

[8] Complex numbers are used in Calculus e.g. Cauchy residue theorem (Theorem for complex integrals) is used to evaluate some difficult real integrals.

[9] Multiplication by $i$ is equivalent to a rotation of $90{}^\circ $ in the argand plane.

[10] A complex number can be thought of as a vector in the Argand plane. This is why complex numbers follow a similar algebra as vector algebra.

[11] Complex numbers are used in AC circuit analysis etc in physics.

Recently Updated Pages

How many sigma and pi bonds are present in HCequiv class 11 chemistry CBSE

Why Are Noble Gases NonReactive class 11 chemistry CBSE

Let X and Y be the sets of all positive divisors of class 11 maths CBSE

Let x and y be 2 real numbers which satisfy the equations class 11 maths CBSE

Let x 4log 2sqrt 9k 1 + 7 and y dfrac132log 2sqrt5 class 11 maths CBSE

Let x22ax+b20 and x22bx+a20 be two equations Then the class 11 maths CBSE

Trending doubts

Which are the Top 10 Largest Countries of the World?

Fill the blanks with the suitable prepositions 1 The class 9 english CBSE

Difference between Prokaryotic cell and Eukaryotic class 11 biology CBSE

Give 10 examples for herbs , shrubs , climbers , creepers

Difference Between Plant Cell and Animal Cell

The Equation xxx + 2 is Satisfied when x is Equal to Class 10 Maths

Change the following sentences into negative and interrogative class 10 english CBSE

Differentiate between homogeneous and heterogeneous class 12 chemistry CBSE

Select the word that is correctly spelled a Twelveth class 10 english CBSE