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Question Number 129807 by bramlexs22 last updated on 19/Jan/21

8.(1+sin (π/8))(1+sin((3π)/8))(1−sin ((5π)/8))(1−sin ((7π)/8))=?

$$\:\mathrm{8}.\left(\mathrm{1}+\mathrm{sin}\:\frac{\pi}{\mathrm{8}}\right)\left(\mathrm{1}+\mathrm{sin}\frac{\mathrm{3}\pi}{\mathrm{8}}\right)\left(\mathrm{1}−\mathrm{sin}\:\frac{\mathrm{5}\pi}{\mathrm{8}}\right)\left(\mathrm{1}−\mathrm{sin}\:\frac{\mathrm{7}\pi}{\mathrm{8}}\right)=?\: \\ $$

Answered by EDWIN88 last updated on 19/Jan/21

sin x = sin (π−x) sin ((3π)/8)=sin ((5π)/8) and sin ((7π)/8)=sin (π/8) ⇔ 8(1+sin (π/8))(1−sin (π/( 8)))(1+sin ((3π)/8))(1−sin ((3π)/8))= ⇔ 8(1−sin^2 (π/( 8)))(1−sin^2 ((3π)/8))= ⇔ 8(cos^2 (π/8))(cos^2 ((3π)/8))= ⇔ 8(((1+cos (π/4))/2))(((1+cos ((3π)/4))/2)) = ⇔ 8(((1+((√2)/2))/2))(((1−((√2)/2))/2))=2(1−(1/2))=1

$$\:\mathrm{sin}\:\mathrm{x}\:=\:\mathrm{sin}\:\left(\pi−\mathrm{x}\right) \\ $$$$\:\mathrm{sin}\:\frac{\mathrm{3}\pi}{\mathrm{8}}=\mathrm{sin}\:\frac{\mathrm{5}\pi}{\mathrm{8}}\:\mathrm{and}\:\mathrm{sin}\:\frac{\mathrm{7}\pi}{\mathrm{8}}=\mathrm{sin}\:\frac{\pi}{\mathrm{8}} \\ $$$$\Leftrightarrow\:\mathrm{8}\left(\mathrm{1}+\mathrm{sin}\:\frac{\pi}{\mathrm{8}}\right)\left(\mathrm{1}−\mathrm{sin}\:\frac{\pi}{\:\mathrm{8}}\right)\left(\mathrm{1}+\mathrm{sin}\:\frac{\mathrm{3}\pi}{\mathrm{8}}\right)\left(\mathrm{1}−\mathrm{sin}\:\frac{\mathrm{3}\pi}{\mathrm{8}}\right)= \\ $$$$\Leftrightarrow\:\mathrm{8}\left(\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} \frac{\pi}{\:\mathrm{8}}\right)\left(\mathrm{1}−\mathrm{sin}\:^{\mathrm{2}} \:\frac{\mathrm{3}\pi}{\mathrm{8}}\right)= \\ $$$$\Leftrightarrow\:\mathrm{8}\left(\mathrm{cos}\:^{\mathrm{2}} \frac{\pi}{\mathrm{8}}\right)\left(\mathrm{cos}\:^{\mathrm{2}} \:\frac{\mathrm{3}\pi}{\mathrm{8}}\right)= \\ $$$$\Leftrightarrow\:\mathrm{8}\left(\frac{\mathrm{1}+\mathrm{cos}\:\frac{\pi}{\mathrm{4}}}{\mathrm{2}}\right)\left(\frac{\mathrm{1}+\mathrm{cos}\:\frac{\mathrm{3}\pi}{\mathrm{4}}}{\mathrm{2}}\right)\:= \\ $$$$\Leftrightarrow\:\mathrm{8}\left(\frac{\mathrm{1}+\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}{\mathrm{2}}\right)\left(\frac{\mathrm{1}−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}}{\mathrm{2}}\right)=\mathrm{2}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\right)=\mathrm{1} \\ $$

Answered by Dwaipayan Shikari last updated on 19/Jan/21

sin(π/8)=sin((7π)/8) sin((5π)/8)=sin((3π)/8) Ψ=8(1−sin^2 (π/8))(1−sin^2 ((3π)/8)) =2(cos(π/4)−cos(π/2))^2 =1

$${sin}\frac{\pi}{\mathrm{8}}={sin}\frac{\mathrm{7}\pi}{\mathrm{8}} \\ $$$${sin}\frac{\mathrm{5}\pi}{\mathrm{8}}={sin}\frac{\mathrm{3}\pi}{\mathrm{8}} \\ $$$$\Psi=\mathrm{8}\left(\mathrm{1}−{sin}^{\mathrm{2}} \frac{\pi}{\mathrm{8}}\right)\left(\mathrm{1}−{sin}^{\mathrm{2}} \frac{\mathrm{3}\pi}{\mathrm{8}}\right) \\ $$$$=\mathrm{2}\left({cos}\frac{\pi}{\mathrm{4}}−{cos}\frac{\pi}{\mathrm{2}}\right)^{\mathrm{2}} =\mathrm{1} \\ $$

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