Question #891ae

1 Answer
Jerald · David B.
Nov 7, 2017

One way is to substitute #45^\circ# for #x#

Explanation:

Let #x = 45^\circ#

#Cos(2*45^\circ)Cos(2*45^\circ)+3sin(2*45^\circ)=2+cos(4*45^\circ)#

#=cos(90^\circ)cos(90^\circ)+3sin(90^\circ)=2+cos(180^\circ)#

#cos(90^\circ)=0#
#sin(90^\circ)=1#
#cos(180^\circ)=0#

#0+3*1=2+0#

#3 \ne 2 #

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