How do you express #sin(pi/ 4 ) * sin( ( 4 pi) / 3 ) # without using products of trigonometric functions?

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How do you express #sin(pi/ 4 ) * sin( ( 4 pi) / 3 ) # without using products of trigonometric functions?

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Answer 1 · 2018-03-13T23:28:25

#-sqrt6/4#

Explanation:

#P = sin (pi/4).sin ((4pi)/3)# =
Trig table and unit circle gives -->
#sin(pi/4) = sqrt2/2#, and #sin ((4pi)/3) = - sin (pi/3) = - sqrt3/2#
Therefor -->
#P = (sqrt2/2)(- sqrt3/2) = - sqrt6/4#

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How do you express #sin(pi/ 4 ) * sin( ( 4 pi) / 3 ) # without using products of trigonometric functions?

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