How do you evaluate #sin(5(pi)/4) - cos(11(pi)/6)#?

1 Answer
Mia
Oct 23, 2016

#-(sqrt2+sqrt3)/2#

Explanation:

In this exercise we will use the following trigonometric identities:

#color(purple)(sin(pi+alpha)=-sinalpha)#
#color(blue)(cos(-alpha)=cosalpha)#

#sin((5pi)/4)-cos((11pi)/6)#
#=sin((4pi+pi)/4)-cos((12pi-pi)/6)#
#=sin((4pi)/4+pi/4)-cos((12pi)/6-pi/6)#
#=sin(pi+pi/4)-cos(2pi-pi/6)#
#=color(purple)(-sin(pi/4)-cos(-pi/6)#
#=color(purple)(-sin(pi/4)-color(blue)(cos(pi/6)#
#=-sin(pi/4)-cos(pi/6)#
#=-sqrt2/2-sqrt3/2#
#=-(sqrt2+sqrt3)/2#

Related questions
See all questions in Trigonometric Functions of Any Angle
Impact of this question
4321 views around the world
You can reuse this answer
Creative Commons License