Skip to content
Question 17 of 17

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.

Yanam BieapBIEAP Intermediate Board 2022Subjective· 7mImportance★★★★★
100% · 17/17 Questions
🔒 Locked · start free trial →

You're viewing a preview — the full solution, concept, methods & PYQ mapping are locked.

Start your 14-day free trial to unlock the full solution →

The expression inside the bracket equals ei 3π/8e^{i\,3\pi/8}, so its 8/38/3-power is eiπ=−1e^{i\pi}=-1.

Let θ=π8\theta=\tfrac{\pi}{8} and put φ=π2−θ=3π8\varphi=\tfrac{\pi}{2}-\theta=\tfrac{3\pi}{8}.

Then sin⁡θ=cos⁡φ\sin\theta=\cos\varphi and cos⁡θ=sin⁡φ\cos\theta=\sin\varphi, so

Numerator =1+cos⁡φ+isin⁡φ=2cos⁡2φ2+2isin⁡φ2cos⁡φ2=2cos⁡φ2 eiφ/2,=1+\cos\varphi+i\sin\varphi=2\cos^2\tfrac{\varphi}{2}+2i\sin\tfrac{\varphi}{2}\cos\tfrac{\varphi}{2}=2\cos\tfrac{\varphi}{2}\,e^{i\varphi/2},

Denominator =1+cos⁡φ−isin⁡φ=2cos⁡φ2 e−iφ/2.=1+\cos\varphi-i\sin\varphi=2\cos\tfrac{\varphi}{2}\,e^{-i\varphi/2}.

…

Unlock everything free for 14 days

  • Full step-by-step solutions
  • Concept-first explanations
  • Methods, shortcuts & mistakes
  • PYQ mapping + timed mock tests

Full access for 14 days. No credit card required.