Question

Find the value of sinπ/14 × sin3π/14 × sin5π/14 × sin7π/14 × sin9π14 × sin11π/14 × sin13π/14 ?

Answer 1

`1/64`
We note that: `color(red)((1)sintheta=cos(pi/2-theta)`
`color(red)((2)sin(pi-theta)=sintheta`
`color(red)((3)cos(-theta)=costheta`
`color(red)((4)2sinthetacostheta=sin(2theta)`

Explanation:

We know that, `sin((7pi)/14)=sin(pi/2)=1` and take,

X=`sin(pi/14)sin((3pi)/14)sin((5pi)/14)sin((9pi)/14)sin((11pi)/14)sin ((13pi)/14)` Using `(1)` above ,we get

`sin((3pi)/14)=cos(pi/2-(3pi)/14)=cos((4pi)/14)=cos((2pi)/7)`

`sin((5pi)/14)=cos(pi/2-(5pi)/14)=cos((2pi)/14)=cos(pi/7)`

`sin((9pi)/14)=cos(pi/2-(9pi)/14)=cos((-2pi)/14)=cos((pi)/7)`

`sin((11pi)/14)=cos(pi/2-(11pi)/14)=cos((-4pi)/14)=cos((2pi)/7)`

`sin((13pi)/14)=sin((14pi-pi)/14)=sin(pi-pi/14)=sin(pi/14)`

So,

`X=sin(pi/14)cos((2pi)/7)cos(pi/7)cos(pi/7)cos((2pi)/7)sin(pi/14)`

`=[sin(pi/14)cos(pi/7)cos((2pi)/7)]^2`

Applying `(4)` several times,we get

`= [1/(2cos(pi/14))xx{2sin(pi/14)cos(pi/14)}cos(pi/7)cos((2pi)/7)]^2`

`=[1/(2cos(pi/14))xx{sin(pi/7)}cos(pi/7)cos((2pi)/7)]^2`

`=[1/(4cos(pi/14))xx{2sin(pi/7)cos(pi/7)}cos((2pi)/7)]^2`

`=[1/(4cos(pi/14))xx{sin((2pi)/7)}cos((2pi)/7)]^2`

`=[1/(8cos(pi/14))xx2sin((2pi)/7)cos((2pi)/7)]^2`

`=[1/(8cos(pi/14))xxsin((4pi)/7)]^2`

`=[1/(8cos(pi/14))xxcos(pi/2-(4pi)/7)]^2`

`=[1/cancel(8cos(pi/14))xxcancelcos((pi)/14)]^2`

`X=1/64`

Answer 2

`rarrsin(pi/14)*sin((3pi)/14)*sin((5pi)/14)*sin((7pi)/14)*sin((9pi)/14)*sin((11pi)/14)*sin((13pi)/14)`

`=1/8[2sin((13pi)/14)*sin(pi/14)][2*sin((11pi)/14)*sin((3pi)/14)][sin(pi/2)][2*sin((9pi)/14)*sin((5pi)/14)`

`=1/8[cos((13pi)/14-pi/14)-cos((13pi)/14+pi/14)][cos((11pi)/14-(3pi)/14)-cos((11pi)/14+(3pi)/14)][cos((9pi)/14-(5pi)/14)-cos((9pi)/14+(5pi)/14)]`

`=1/8[1+cos((6pi)/7)][1+cos((4pi)/7)][1+cos((2pi)/7)]`

`=1/8[2cos^2((3pi)/7)][2cos^2((2pi)/7)][2cos^2((pi)/7)]`

`=[cos(pi/7)*cos((2pi)/7)*cos((3pi)/7)]^2`

`=[1/(2sin(pi/7)){2sin(pi/7)cos(pi/7)*cos((2pi)/7)*cos((3pi)/7)}]^2`

`=[1/(2*2sin(pi/7)){2*sin((2pi)/7)*cos((2pi)/7)*cos((3pi)/7)}]^2`

`=[1/(4*2sin(pi/7)){2*sin((4pi)/7)*cos((3pi)/7)}]^2`

`=[1/(8sin(pi-(6pi)/7)){2*sin(pi-(3pi)/7)*cos((3pi)/7)}]^2`

`=[1/(8sin((6pi)/7)){2*sin((3pi)/7)*cos((3pi)/7)}]^2`

`=[1/(8sin((6pi)/7))*sin((6pi)/7)]^2=[1/8]^2=1/64`