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

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

God bless America...

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