How do you express #cos(pi/ 3 ) * sin( ( 5 pi) / 6 ) # without using products of trigonometric functions?

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How do you express #cos(pi/ 3 ) * sin( ( 5 pi) / 6 ) # without using products of trigonometric functions?

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Answer 1 · 2016-05-26T19:12:50

#(1/2)sin (pi/6)#

Explanation:

#P = cos (pi/3).sin ((5pi)/6)#
Trig table -->
#cos (pi/3) = 1/2#
Unit circle, property of supplementary arcs -->
#sin ((5pi)/6) = sin (pi - pi/6) = sin (pi/6)#
P can be expressed as:
#P = (1/2)sin (pi/6)#

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How do you express #cos(pi/ 3 ) * sin( ( 5 pi) / 6 ) # without using products of trigonometric functions?

1 Answer

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