Narendra Beniwal
Last Activity: 5 Years ago
cos 5pi/8=cos(pi/2+pi/8)=-sin pi/8
cos 3pi/8=cos(pi/2-pi/8)=sin pi/8
cos 7pi/8=cos(pi-pi/8)=-cos pi/8. Now the expression (1+cos pi/8)(1+cos 3pi/8)(1+cos 5pi/8)(1+cos 7pi/8) can be written as,
(1+cos pi/8)(1+sin pi/8)(1-sin pi/8)(1-cos pi/8)
=[1-(cos pi/8)^2][1–(sin pi/8)^2]
=(sin pi/8)^2×(cos pi/8)^2
=1/4×(2 sin pi/8.cos pi/8)^2
=1/4(sin 2x pi/8)^2
=1/4(sin pi/4)^2
=1/4×1/[(2)^1/2]^2
=1/4×1/2
=1/8