Question

cos 20 cos 40 cos 60 cos 80 = 1/16​

Answer 1

Answer:

Hii

Step-by-step explanation:

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

Answer 2

${\bold{\underline{\boxed{Solution}}}}$

Given : cos20 cos40 cos60 cos80= $\large\tt\dfrac{1}{2}$

Taking L.H.S

cos20 cos40 cos60 cos80

= cos60 (cos20 cos40)cos80

Multiplying and Divide by 2

= $\large\tt\dfrac{1}{2}\times\large\tt\dfrac{1}{2}$ (2 cos20 cos40)cos80

=$\large\tt\dfrac{1}{4}$ [cos(40+20)+cos(40−20)cos80]

[As 2cosA cosB = cos(A+B)+cos(A−B)]

=$\large\tt\dfrac{1}{4}$[(cos60+cos20)cos80]

=$\large\tt\dfrac{1}{4}$(12+cos20)cos80]

=$\large\tt\dfrac{1}{4}$[12cos80+cos80 cos20]

Again Multiplying and Divide by 2 we get

= $\large\tt\dfrac{1}{4}\times\large\tt\dfrac{1}{2}$[cos80+cos(80+20)+cos(80−20)]

=$\large\tt\dfrac{1}{8}$[cos80+cos(80+20)+cos(80−20)]

=$\large\tt\dfrac{1}{8}$[cos80+cos100+cos60]

=$\large\tt\dfrac{1}{8}$[cos80+cos(180−80)+cos60]

=$\large\tt\dfrac{1}{8}$[cos80−cos80+cos60]

=$\large\tt\dfrac{1}{8}$ × cos60

$\longrightarrow$ $\large\tt\dfrac{1}{8}\times\large\tt\dfrac{1}{2}$ = $\large\tt\dfrac{1}{16}$ = R.H.S

Hence Verified.