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

2 Answers
Apr 17, 2018

#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#

Apr 17, 2018

#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#