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Question: $\sin^4 \frac{\pi}{8} + \sin^4 \frac{3\pi}{8} + \sin^4 \frac{5\pi}{8} + \sin^4 \frac{7\pi}{8} = \fra...

sin⁡4π8+sin⁡43π8+sin⁡45π8+sin⁡47π8=23\sin^4 \frac{\pi}{8} + \sin^4 \frac{3\pi}{8} + \sin^4 \frac{5\pi}{8} + \sin^4 \frac{7\pi}{8} = \frac{2}{3}

Answer

False

Explanation

Solution

Let S=sin⁡4π8+sin⁡43π8+sin⁡45π8+sin⁡47π8S = \sin^4 \frac{\pi}{8} + \sin^4 \frac{3\pi}{8} + \sin^4 \frac{5\pi}{8} + \sin^4 \frac{7\pi}{8}. Using sin⁡(π−θ)=sin⁡(θ)\sin(\pi - \theta) = \sin(\theta), we have sin⁡5π8=sin⁡3π8\sin \frac{5\pi}{8} = \sin \frac{3\pi}{8} and sin⁡7π8=sin⁡π8\sin \frac{7\pi}{8} = \sin \frac{\pi}{8}. So, S=2(sin⁡4π8+sin⁡43π8)S = 2 (\sin^4 \frac{\pi}{8} + \sin^4 \frac{3\pi}{8}). Using sin⁡(π2−θ)=cos⁡(θ)\sin(\frac{\pi}{2} - \theta) = \cos(\theta), we have sin⁡3π8=cos⁡π8\sin \frac{3\pi}{8} = \cos \frac{\pi}{8}. So, S=2(sin⁡4π8+cos⁡4π8)S = 2 (\sin^4 \frac{\pi}{8} + \cos^4 \frac{\pi}{8}). Using sin⁡4θ+cos⁡4θ=1−12sin⁡2(2θ)\sin^4 \theta + \cos^4 \theta = 1 - \frac{1}{2}\sin^2(2\theta), with θ=π8\theta = \frac{\pi}{8}, 2θ=π42\theta = \frac{\pi}{4}. S=2(1−12sin⁡2(π4))=2(1−12(12)2)=2(1−12×12)=2(1−14)=2×34=32S = 2 (1 - \frac{1}{2}\sin^2(\frac{\pi}{4})) = 2 (1 - \frac{1}{2}(\frac{1}{\sqrt{2}})^2) = 2 (1 - \frac{1}{2} \times \frac{1}{2}) = 2 (1 - \frac{1}{4}) = 2 \times \frac{3}{4} = \frac{3}{2}. Since 32≠23\frac{3}{2} \neq \frac{2}{3}, the statement is false.