How do you express #sin(pi/ 6 ) * cos( ( 15 pi) / 12 ) # without using products of trigonometric functions?

2 Answers
Nghi N.
Mar 26, 2016

#-sqrt2/4#

Explanation:

#P = sin (pi/6)cos ((15pi)/12).#
Trig table --> #sin pi/6 = 1/2#
#cos ((15pi)/12) = cos ((-9pi)/12 + 2pi) = cos ((-3pi)/4) =#
#= cos ((3pi)/4) = -sqrt2/2.#
#P = (1/2)(-sqrt2/2) = -sqrt2/4#

Bdub
Apr 4, 2016

#sin (pi/6) cos ((15pi)/12)=1/2 sin ((17pi)/12) - 1/2sin ((13pi)/12)#

Explanation:

#2 sin A cos B=sin (A+B) + sin (A-B) #
#sin A cos B=1/2 (sin (A+B) + sin (A-B))#
#A=pi/6, B=(15pi)/12#
#sin (pi/6) cos ((15pi)/12)=1/2 (sin (pi/6+(15pi)/12) + sin (pi/6-(15pi)/12))#
#=1/2 (sin ((17pi)/12) + sin (-(13pi)/12))#
#=1/2 sin ((17pi)/12) - 1/2sin ((13pi)/12)#

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