# Mathematical IdentityProve the below identity: sin20*sin40*sin60*sin80=3/16

sciencesolve | Certified Educator

You need to remember the next trigonometric identity, such that:

`cos alpha - cos beta = 2sin((alpha + beta)/2)sin((beta - alpha)/2)`

Considering the product `sin 20^o*sin 80^o` yields:

`sin 20^o*sin 80^o = (cos alpha - cos beta)/2`

Notice that `alpha + beta = 20^o` and `-alpha + beta = 80^o` , hence, you need to solve for alpha and beta the following system of equations, such that:

`{(alpha + beta = 20^o),(-alpha + beta = 80^o):} => 2 beta = 100^o=> beta = 50^o => alpha = -30^o`

`sin 20^o*sin 80^o = (cos (-30^o) - cos 50^o)/2`

`Since cos(-alpha) = cos alpha => cos (-30^o) = cos 30^o`

`sin 20^o*sin 80^o = (cos 30^o - cos 50^o)/2`

You need to consider the product `sin 40^o*sin 60^o` such that:

`sin 40^o*sin 60^o = (cos alpha - cos beta)/2`

`{(alpha + beta = 40^o),(-alpha + beta = 60^o):} => 2 beta = 100^o => beta = 50^o => alpha = -10^o`

`sin 40^o*sin 60^o = (cos 10^o - cos 50^o)/2`

Hence, you may write the given product such that:

`sin 20^o*sin 40^o*sin 60^o*sin 80^o = (cos 30^o - cos 50^o)/2*(cos 10^o - cos 50^o)/2`

You need to test if the expression below holds:

`(cos 30^o - cos 50^o)(cos 10^o - cos 50^o)/4 = 3/16`

`(cos 30^o - cos 50^o)(cos 10^o - cos 50^o) = 3/4`

`cos 30^o*cos 10^o - cos 30^o*cos 50^o - cos 50^o*cos 10^o + cos 50^o*cos50^o = 3/4`

`sqrt3/2(cos 10^o - cos 50^o) + cos 50^o(cos 10^o - cos 50^o) = 3/4`

`cos 10^o - cos 50^o = 2 sin 30^o*sin 20^o = sin 20^o`

`sin 20^o*(sqrt3)/2 + sin 20^o*cos 50^o`

`0.342*(sqrt3)/2 +0.342*0.642 ~~ 3/4`

Hence, testing if the equality holds yields that` sin 20^o*sin 40^o*sin 60^o*sin 80^o = 3/16.`

giorgiana1976 | Student

In order to check up the true value of the above identity, we have to use the following formula:

sin (a)*sin(b) = [cos (a-b) - cos (a+b)]/2

sin20*sin40 = [cos (20-40) - cos (20+40)]/2

sin20*sin40 = [cos (-20) - cos (60)]/2

But cos (-a) = cos (a)

sin20*sin40 = [cos (20) - cos (60)]/2

sin60*sin80 = [cos (60-80) - cos (60+80)]/2

sin60*sin80 = [cos (20) - cos (140)]/2

But - cos (140) = cos (40)

sin20*sin40*sin60*sin80 = {[cos (20) - cos (60)]/2}*{[cos (20) + cos (40)]/2}

{[cos (20) - cos (60)]/2}*{[cos (20) + cos (40)]/2} =

=[cos^2 (20) - cos (20)*cos (60) + cos (20)*cos (40) - cos (40)*cos (60)]/4= [1 - cos (20) + cos (40) + cos (20) +

cos (60) - cos (40)]/8= [1+cos (60)]/8=[1+1/2]/8=3/16