How do you express #sin(pi/ 4 ) * cos( ( 5 pi) / 4 ) # without using products of trigonometric functions?

2 Answers
Nghi N.
Apr 4, 2016
  • 1/2

Explanation:

#P = sin (pi/4)cos ((5pi)/4)#
Trig table --> #sin (pi/4) = sqrt2/2.#
#cos ((5pi)/4 = cos (pi/4 + pi) = -cos (pi/4) = - sqrt2/2#
#P = (-sqrt2/2)(sqrt2/2) = - 1/2#

Bdub
Apr 4, 2016

#sin (pi/4) cos ((5pi)/4)=1/2 sin ((3pi)/2) -1/2 sin (pi) =-1/2#

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/4, B= (5pi)/4#

#sin (pi/4) cos ((5pi)/4)=1/2 (sin (pi/4+(5pi)/4) + sin (pi/4-(5pi)/4))#

#=1/2 (sin ((6pi)/4) + sin ((-4pi)/4))#

#=1/2 (sin ((3pi)/2) + sin (-pi))#

#=1/2 (sin ((3pi)/2) -sin (pi))#

#=1/2 sin ((3pi)/2) -1/2 sin (pi)#

#=1/2 *(-1)-1/2 (0)#

#= -1/2#

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