If #alpha = pi/13# then the value of #prod_(r=1)^6 cos(ralpha)# is equal to ?

Question

If #alpha = pi/13# then the value of #prod_(r=1)^6 cos(ralpha)# is equal to :

(A)#1/64#

(B)#-1/64#

(C)#1/32#

(D)#-1/8#

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Answer 1 · 2018-06-28T12:56:03

Given #alpha=pi/13=>13alpha=pi#

Let

#P=prod_(r=1)^6 cos(ralpha)#

#=>P sinalpha=1/2*2sinalphacosalphaprod_(r=2)^6 cos(ralpha)#

#=1/4*2sin2alphacos2alphaprod_(r=3)^6 cos(ralpha)#

#=1/8*2sin4alphacos4alpha*cos3alpha *cos5alpha*cos6alpha#

#=1/8*sin8alpha*cos3alpha *cos5alpha*cos6alpha#

#=1/8*sin(13alpha-5alpha)*cos3alpha *cos5alpha*cos6alpha#
#=1/8*sin(pi-5alpha)*cos3alpha *cos5alpha*cos6alpha#

#=1/16*2sin5alphacos5alpha *cos3alpha*cos6alpha#

#=1/16*2sin5alphacos5alpha *cos3alpha*cos6alpha#
#=1/16sin10alpha *cos3alpha*cos6alpha#

#=1/16sin(13alpha -3alpha)cos3alpha*cos6alpha#

#=1/32*2sin(pi -3alpha)cos3alpha*cos6alpha#

#=1/32*2sin 3alphacos3alpha*cos6alpha#

#=1/64*2sin 6alphacos6alpha#

#=1/64*sin 12alpha#

#=>P sinalpha=1/64*sin (13alpha-alpha)#

#=>P sinalpha=1/64*sin (pi-alpha)#
#=>P sinalpha=1/64*sin alpha#
#=>P=1/64#