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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))=?

8.(1+sinπ8)(1+sin3π8)(1−sin5π8)(1−sin7π8)=?

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

sinx=sin(π−x)sin3π8=sin5π8andsin7π8=sinπ8⇔8(1+sinπ8)(1−sinπ8)(1+sin3π8)(1−sin3π8)=⇔8(1−sin2π8)(1−sin23π8)=⇔8(cos2π8)(cos23π8)=⇔8(1+cosπ42)(1+cos3π42)=⇔8(1+222)(1−222)=2(1−12)=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

sinπ8=sin7π8sin5π8=sin3π8Ψ=8(1−sin2π8)(1−sin23π8)=2(cosπ4−cosπ2)2=1

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