Q 17 Let K = sin10 sin20 sin30 sin80 then 256 K = - Maths - Straight Lines

Question

Q 17 Let K = sin 10 ° sin 20 ° sin 30 ° sin 80 °
then 256 K =
(a) 1
(b) 2
(c) 3
(d) 13

Lovina Kansal · Accepted Answer

Dear student
Consider,sin10 sin50 sin70=sin(60-10)sin(60+10)sin10=sin260-sin210sin10   ∵sin2A-sin2B=sinA+BsinA-B=322-sin210sin10=34-sin210sin10=143-4sin210sin10=143sin10-4sin310=14sin3×10    ∵sin3A=3sinA-4sin3A=14sin30=14×12=18So, sin10 sin50 sin70=18Consider,sin20 sin40 sin80=sin20sin60-20sin60+20=sin20sin260-sin220=sin20322-sin220=sin2034-sin220=143sin20-4sin320=14sin3×20=14sin60=14×32=38So, sin20 sin40 sin80=38Consider,K=sin10 sin20 sin30 sin40 sin50 sin60 sin70 sin80=sin10 sin50 sin70sin20 sin40 sin80sin30 sin60=18×38×12×32=3256K=sin10 sin20 sin30 sin40 sin50 sin60 sin70 sin80=3256So,256K=256×3256=3Hence,256K=3
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Q.17. Let K = sin 10 ° sin 20 ° sin 30 ° . . . . . sin 80 ° then 256 K = (a) 1 (b) 2 (c) 3 (d) 13

Q.17. Let K = $\sin 10 ° \sin 20 ° \sin 30 ° . . . . . \sin 80 °$ then 256 K =
(a) 1
(b) 2
(c) 3
(d) 13