Verify if cos3a=(x^6+1)/2x^3 cosa=(x^2+1)/2x

cos a = (x^2 + 1)/2x

let
usdetermine cos 3a:

We know
that:

Let us rewrite cos 3a
:

cos 3a = cos (2a + a)

But we
know that :

cos(a+b) = cosa*cosb -
sina*sinb

==> cos(2a + a) = cos2a*cosa -
sin2a*sina

Now we know that
:

cos2a = cos^2 a - sin^2
a

sin2a = 2sina*cosa

=>
cos(2a+a) = (cos^2 a - sin^2 a)*cosa - 2cosa*sin^2
a

                      = cos^3 a - cosa*sin^2 a -
2cosa*sin^2 a

                        = cos^3 a -
3cosa*sin^2 a

But sin^2 a = 1-cos^2
a

==> cos(2a+a) = cos^3 a - 3cosa(1-cos^2
a)

                         = cos^3 a - 3cosa + 3cos^3
a

                         = 4cos^3 a - 3cos
a

Now substitute with cosa = (x^2 +
1)/2x

==> cos 3a = 4[ (x^2 + 1)/2x]^3 - 3 (x^2 +
1)/2x

                    = (x^2 + 1)^3/ 2x^3 - 3(x^2
+1)/2x

                     = [(x^2 + 1)^3 - 3x^2(x^2
+1)]/2x^3

                     = (x^2 + 1)*[ (x^2 + 1)^2 -
3x^2]/2x^3

                     = (x^2 +1)*( x^4 + 2x^2 + 1
- 3x^2)/2x^3

                     = (x^2 + 1) *(x^4 - x^2 +
1) /2x^3

                      = (x^6 - x^4 + x^2 + x^4 -
x^2 +1)/2x^3

                     = (x^6
+1)/2x^3

==> cos3a = (x^6 +
1)/2x^3