Question

cos 20 cos40 cos 60 cos 80=1/16



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Answer 1

cos20 cos40 cos60 cos80= 1/16

l.h.s. :

cos20 cos40 1/2 cos80 (cos60 = 1/2)

multiply nd divide by 2

cos20 cos40 1/2 cos80 (cos60 = 1/2)

multiply nd divide by 2

1/4 (2 cos20 cos40 cos80)

1/4 (cos(20+80)+ cos(20-80)) cos40 (2cosa cosb= cos(a+b) + cos(a-b))

1/4 (cos(-60)+ cos(100)) cos40

1/4(1/2 + cos100)cos40

1/8 cos40+ 1/4 (cos40 cos100)

multiplt nd divide by 2

2/2(1/8 cos40) + 1/8(2 cos40 cos100)

1/8 cos40+ 1/8 (cos140+ cos(-60)) (2cosa cosb= cos(a+b) cos(a-b))

1/8 cos40+ 1/8 cos140 + 1/16 (cos60= 1/2)

1/8(cos40+cos140) + 1/16

1/8(2 cos90 cos(-50)) + 1/16 (as above identity)

cos90= 0

1/16

= r.h.s

hence proved........

Answer 2

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$\it\huge\mathfrak\red{Hello}$]

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L.H.S.

=(cos20°.cos40°)cos60°.cos80°

=1/2[cos(20° + 40°) + cos(20° – 40°)]×1/2×cos80°

=1/4[cos60° + cos(-20°)]cos80°

=1/4[cos60°cos80° + cos20°cos80°]

=1/4[1/2cos80° + 1/2{cos(20° + 80°) + cos(20° – 80°)}]

=1/8[cos80° + {cos100° + cos(-60°)}]

=1/8[cos80° + cos100° + cos60°]

=1/8[cos80° +cos(180° – 80°) +cos60°]

=1/8[cos80° – cos80° + cos60°]

=1/8 ×cos60°

=1/8 × 1/2

=1/16 = R.H.S

L.H.S = R.H.S = 1/16 Hence proved

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$\it\huge\mathfrak\red{thanks}$]

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