Question #fdeca

1 Answer
Oct 26, 2014

Answer:

#cos(x)#

Explanation:

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)#