Question

Prove that sinθ.sin(60 - θ).sin(60 + θ) = 1/4 sin 3θ .

Answer 1

use d identity 2sinAsinB=cos(A-B)-cos(A+B)

Answer 2

use it where ?

Answer 3

or use d identity sin(a+b) *Sin(a-b)= sin^2a- Sin^2b

Answer 4

used the 1st identity for last two terms i.e. Sin(60+x)*Sin(60-x)

Answer 5

1st multiply d num n denom by 2

Answer 6

is there a dr ?

Answer 7

then 2Sin(60+x)*Sin(60-x)=cos2x-cos120

Answer 8

then multiply it by sinx

Answer 9

u'll have (sinxcos2x-sinxcos120)/2

Answer 10

den cos2x=1-2sin^2x and simplify

Answer 11

and then .. i don't know akash after

Answer 12

and is that correct ?

Answer 13

yes..

Answer 14

and then what to do ?

Answer 15

use co120=-1/2 and sin3x=3sinx-4sin^3x

Answer 16

cos120=-1/2...just u need to simplify and arrange the terms

Answer 17

try it....

Answer 18

yea .. okay got it , thanks @akash123 :)

Answer 19

good...:)