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Q.Show that one value of (1+sin⁡π8+icos⁡π81+sin⁡π8−icos⁡π8)8/3\left(\dfrac{1+\sin\frac{\pi}{8}+i\cos\frac{\pi}{8}}{1+\sin\frac{\pi}{8}-i\cos\frac{\pi}{8}}\right)^{8/3} is −1-1.

Telangana TsbieTelangana Board of Intermediate Education 2024Subjective· 7mImportance★★★★★
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Rewrite sin⁡θ,cos⁡θ\sin\theta,\cos\theta using θ=π8\theta=\tfrac{\pi}{8} in terms of a complementary angle ψ=π2−θ\psi=\tfrac{\pi}{2}-\theta, reducing numerator and denominator to conjugate exponentials.

Let θ=π8\theta=\dfrac{\pi}{8} and ψ=π2−θ=3π8\psi=\dfrac{\pi}{2}-\theta=\dfrac{3\pi}{8}. Since sin⁡θ=cos⁡ψ\sin\theta=\cos\psi and cos⁡θ=sin⁡ψ\cos\theta=\sin\psi:

Numerator =1+sin⁡θ+icos⁡θ=1+cos⁡ψ+isin⁡ψ=1+\sin\theta+i\cos\theta = 1+\cos\psi+i\sin\psi

Using 1+cos⁡ψ=2cos⁡2ψ21+\cos\psi=2\cos^2\tfrac{\psi}{2} and sin⁡ψ=2sin⁡ψ2cos⁡ψ2\sin\psi=2\sin\tfrac{\psi}{2}\cos\tfrac{\psi}{2}:

=2cos⁡2ψ2+i⋅2sin⁡ψ2cos⁡ψ2=2cos⁡ψ2(cos⁡ψ2+isin⁡ψ2)=2cos⁡ψ2 eiψ/2= 2\cos^2\tfrac{\psi}{2} + i\cdot2\sin\tfrac{\psi}{2}\cos\tfrac{\psi}{2} = 2\cos\tfrac{\psi}{2}\left(\cos\tfrac{\psi}{2}+i\sin\tfrac{\psi}{2}\right) = 2\cos\tfrac{\psi}{2}\,e^{i\psi/2}

Similarly, Denominator =1+cos⁡ψ−isin⁡ψ=2cos⁡ψ2 e−iψ/2=1+\cos\psi-i\sin\psi = 2\cos\tfrac{\psi}{2}\,e^{-i\psi/2}

So the ratio is:

2cos⁡ψ2 eiψ/22cos⁡ψ2 e−iψ/2=eiψ=cos⁡ψ+isin⁡ψ\dfrac{2\cos\frac{\psi}{2}\,e^{i\psi/2}}{2\cos\frac{\psi}{2}\,e^{-i\psi/2}} = e^{i\psi} = \cos\psi+i\sin\psi, with ψ=3π8\psi=\dfrac{3\pi}{8}

Raising to the power 83\tfrac{8}{3} (taking the principal value):

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