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

1 Answer

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

Explanation:

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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