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

hala718 | Certified Educator

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

neela | Student

To verify cos3a = (x^6+1)/2x^3, cosa = (x^2+1)/2x.

Solution:

We know cos3a = cos2a*cosa - sin2asina = (2cos^2a-1)cosa - (2sin^2acosa) = 2cos^3a -cosa - 2(1-cos^2a)cosa  =  4cos^3a - 3cosa.

cos^3a = 4cos^3a - 3cosa. Substitute the value cosa = (x^2+1)/2x.

RHS 4cos3^a - cosa = 4{(x^2+1)/2x}^3 -3 (x^2+1)/2x)

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

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

{(x^2+1)/2x}(1/2x)^2{4x^4+8x^2+4- 12x^2}

= (x^2+1)(1/8x^3){ 4x^4-4x^2+4}

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

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

= (1/2x^3) {( x^2)^3+(1)^3}, as (a+1)(a^2-a+1) = a^3+1

= (x^6+1)/2x^3 which is equal to cos^3a.

Therefore  if cosa = (x^2+1)/2x , then cos3a = (x^6+1)/2x^3 is correct.

giorgiana1976 | Student

For the beginning, we'll write the formula for cos 3a:

cos 3a = cos (2a+a) = cos 2a*cos a - sin 2a*sin a

cos 3a = 4(cos a)^3 - 3 cos a

We'll re-write the expression that has to be verified:

2 cos 3a = x^6/x^3 + 1/x^3

We'll reduce like terms:

2 cos 3a = x^3 + 1/x^3

Now, we'll substitute the formula into the expression:

2 cos 3a = 2*[ 4(cos a)^3 - 3 cos a] = 8(cos a)^3 - 6 cos a

We could write 8(cos a)^3 = (2*cos a)^3

But 2cos a = x^2/x + 1/x

2cos a = x + 1/x

We'll substitute 2cos a = x + 1/x into the expression:

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

We'll expand the cube:

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

We'll elimiknate like terms:

2 cos 3a = (x + 1/x)^3 = x^3 + (1/x)^3

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

cos 3a = (x^6+1)/2x^3