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

1 Answer
Nghi N.
Feb 9, 2016

#P= - [(sqrt3)(sqrt(2 - sqrt2))]/4#

Explanation:

Product: #P = sin (pi/8).cos ((5pi)/6#).
Find separately #sin (pi/8)# and #cos ((5pi)/6).#
a. Find #sin (pi/8)#. Use trig identity: cos 2x = 1 - 2sin^2 x.
#cos (pi/4) = sqrt2/2 = 1 - 2sin^2 (pi/8)#
#2sin^2 (pi/8) = 1 - sqrt2/2 = (2 - sqrt2)/2#
#sin^2 (pi/8) = (2 - sqrt2)/4#
#sin (pi/8) = sqrt(2 - sqrt2)/2# (#sin pi/8# is positive).
b. Find #cos ((5pi)/6)#
#cos ((5pi)/6) = cos (-pi/6 + pi) = -cos pi/6 = -sqrt3/2#

Finally, #P = -[(sqrt3)(sqrt(2 - sqrt2))]/4#

Related questions
See all questions in Products, Sums, Linear Combinations, and Applications
Impact of this question
1063 views around the world
You can reuse this answer
Creative Commons License