How do you express $sin(pi/ 6 ) * cos( ( 5 pi) / 8 )$ without using products of trigonometric functions?

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How do you express $sin(pi/ 6 ) * cos( ( 5 pi) / 8 )$ without using products of trigonometric functions?

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Answer 1 · 2016-03-22T04:01:13

$P = -(1/2)sin (pi/8)$

Explanation:

$P = sin (pi)/6.cos ((5pi)/8)$
Trig table --> sin (pi/6) = 1/2
Trig unit circle -->
$cos((5pi)/8) = cos (pi/8 + (4pi)/8) = cos (pi/8) + pi/2) = - sin (pi/8)$
(Complementary arcs)
#P = - (1/2)sin (pi/8)

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How do you express $sin(pi/ 6 ) * cos( ( 5 pi) / 8 )$ without using products of trigonometric functions?

1 Answer

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How do you express sin(pi/ 6 ) * cos( ( 5 pi) / 8 ) sin(pi/ 6 ) * cos( ( 5 pi) / 8 ) without using products of trigonometric functions?

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How do you express $sin(pi/ 6 ) * cos( ( 5 pi) / 8 )$ without using products of trigonometric functions?