How do you express #cos(pi/ 3 ) * sin( ( 7 pi) / 8 ) # without using products of trigonometric functions?

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Answer 1 · 2016-02-13T02:31:08

Use Sum or Difference
#cos(pi/3)*sin((7pi)/8)=1/2*sin((13pi)/24)-1/2*sin ((5pi)/24)#

This is the derivation

#sin (x+y)=sin x cos y+ cos x sin y" " " "#1st equation
#sin (x-y)=sin x cos y- cos x sin y" " " "#2nd equation

subtract 2nd from the 1st

#sin (x+y)-sin(x-y)=2*cos x sin y#
it follows

#cos x sin y=1/2[sin (x+y)-sin(x-y)]#

Now, let #x=pi/3# and #y=(7pi)/8#

#cos (pi/3) sin ((7pi)/8)=1/2[sin (pi/3+(7pi)/8)-sin(pi/3-(7pi)/8)]#

#cos (pi/3) sin ((7pi)/8)=1/2[sin ((8pi+21pi)/24)-sin ((8pi-21pi)/24)]#

#cos (pi/3) sin ((7pi)/8)=1/2[sin ((29pi)/24)-sin ((-13pi)/24)]#

#cos (pi/3) sin ((7pi)/8)=1/2[sin (pi+(5pi)/24)-sin ((-13pi)/24)]#

Note: #sin ((-13pi)/24)=-sin ((13pi)/24)#

so that

#cos (pi/3) sin ((7pi)/8)=1/2[sin (pi+(5pi)/24)-(-sin ((13pi)/24))]#

#cos (pi/3) sin ((7pi)/8)=1/2[sin (pi+(5pi)/24)+sin ((13pi)/24)]#

#cos (pi/3) sin ((7pi)/8)=#
#1/2[sin pi cos((5pi)/24)+cos pi sin ((5pi)/24)+sin ((13pi)/24)]#

#cos (pi/3) sin ((7pi)/8)=1/2[sin((13pi)/24)-sin ((5pi)/24)]#

final answer

#1/2*sin((13pi)/24)-1/2*sin ((5pi)/24)#

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