Recall our trig identities:
#sin(2x) = 2sin(x)cos(x)#
#cos(2x) = cos^2(x) – sin^2(x) #
#sin^2(x)+cos^2(x)=1#
Looking at our problem:
#cos(2x) cos(x) + sin(2x) sin(x)#
Let's substitute for #sin(2x)#. This will give us:
#cos(2x) cos(x) + (2sin(x)cos(x))* sin(x)#
#=cos(2x)cos(x)+2sin^2(x)cos(x)#
Now, let's substitute #sin^2(x)# using the identity #sin^2(x)+cos^2(x)=1#
This gives us:
#cos(2x)cos(x)+2sin^2(x)cos(x)#
#= cos(2x)cos(x)+2(1-cos^2(x))*cos(x)#
Now, let's substitute #cos(2x)# using the identity #cos(2x) = cos^2(x) – sin^2(x) #
This gives us:
#cos(2x)cos(x)+2(1-cos^2(x))*cos(x)#
#=(cos^2(x)-sin^2(x))*cos(x)+2(1-cos^2(x))*cos(x)#
Now, let's substitute the sin^2(x) inside the parenthesis using the identity, #sin^2(x)+cos^2(x)=1#
This gives us:
#(cos^2(x)-sin^2(x))*cos(x)+2(1-cos^2(x))*cos(x)#
#=(cos^2(x)-(1-cos^2(x)))*cos(x)+2(1-cos^2(x))*cos(x)#
#=(cos^2(x)-1+cos^2(x))*cos(x)+2(1-cos^2(x))*cos(x)#
#=(2cos^2(x)-1)*cos(x)+2(1-cos^2(x))*cos(x)#
Now, multiplying through with our #cos(x)#'s:
#(2cos^2(x)-1)*cos(x)+2(1-cos^2(x))*cos(x)#
#=2cos^3(x)-cos(x)+2(cos(x)-cos^3(x))#
#=2cos^3(x)-cos(x)+2cos(x)-2cos^3(x)#
The #2cos^3(x)# and the #-2cos^3(x)# cancel out,
and we are left with:
#-cos(x)+2cos(x)#
which equals:
#cos(x)#
