Using the identity cos(A+B) = cosAcosB - sinAsinBcos(A+B)=cosAcosBsinAsinB, how do you prove that 1/4cos(3A) = cos^3A - 3/4cosA14cos(3A)=cos3A34cosA?

3 Answers
Jun 12, 2016

I'm going to work from the right hand side. For cos3Acos3A, we know that:

color(green)(cos3A = cos(A + 2A) = cosAcolor(highlight)(cos2A) - sinAcolor(highlight)(sin2A)).cos3A=cos(A+2A)=cosAcos2AsinAsin2A.

Next, recall that:

  • color(highlight)(sin2A) = sinAcosA + cosAsinA = color(highlight)(2sinAcosA)sin2A=sinAcosA+cosAsinA=2sinAcosA
  • color(highlight)(cos2A) = cosAcosA - sinAsinA = color(highlight)(cos^2A - sin^2A)cos2A=cosAcosAsinAsinA=cos2Asin2A

So, we now have:

cos3A = cosA(color(highlight)(cos^2A - sin^2A)) - sinA(color(highlight)(2sinAcosA))cos3A=cosA(cos2Asin2A)sinA(2sinAcosA)

= cos^3A - sin^2AcosA - 2sin^2AcosA=cos3Asin2AcosA2sin2AcosA

= cos^3A - 3color(red)(sin^2)AcosA=cos3A3sin2AcosA

What I'm shooting for at this point is the result you get on the left hand side when you multiply the right hand side by 44: color(purple)(4cos^3A - 3cosA)4cos3A3cosA. That requires all functions to be coscos here.

So, we have to reference the identity that you definitely know: color(red)(sin^2A + cos^2A = 1)sin2A+cos2A=1. This gives us:

= cos^3A - 3(color(red)(1 - cos^2A))cosA=cos3A3(1cos2A)cosA

= cos^3A - 3(cosA - cos^3A)=cos3A3(cosAcos3A)

= cos^3A - 3cosA + 3cos^3A=cos3A3cosA+3cos3A

= color(purple)(4cos^3A - 3cosA)=4cos3A3cosA

Almost there! Finally, since this was equal to cos3Acos3A, we have to divide by 44 to get:

1/4cos3A = 1/4*[4cos^3A - 3cosA]14cos3A=14[4cos3A3cosA]

=> color(blue)(1/4cos3A = cos^3A - 3/4cosA)14cos3A=cos3A34cosA

Jun 13, 2016

Using only the given identity of cos(A+B)cos(A+B) the other given identity can be proved as follows.

Explanation:

General Discussion

The identity to be assumed is

cos(A+B)=cosAcosB-sinAsinB...(1)

As this is valid for all real values of A and B , we can put convenient values to get different identities as follows.

Method - I( A Tricky one)

Putting B= -B in equation (1)

cos(A-B)=cosAcos(-B)-sinAsin(-B)

=>cos(A-B)=cosAcosB+sinAsinB…(1.1)

Adding (1) and (1.1) we get

cos(A+B)+cos(A-B)=2cosAcosB…(1.2)

Now Putting B=2A in (1.2)

cos(A+2A)+cos(A-2A)=2cosAcos2A

cos3A+cosA=2cosAcos2A

cos3A=cosA(2cos2A-1)…(1.3)

Putting B=A in (1.2)

cos(A+A)+cos(A-A)=2cosAcosA

=>cos2A+cos0=2cos^2A

cos2A=2cos^2A-1…(1.4)

Finally combining (1.4) with (1.3) we get

cos3A=cosA(2(2cos^2A-1)-1)

=>cos3A=4cos^3A-3cosA

Re-arrenging and deviding bothsides by 4
cos^3A-3/4cosA=1/4cos3A
Proved

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Method-II (A long one)

Putting B=A in equation(1)

cos(A+A)=cosAcosA-sinAsinA

cos2A=cos^2A-sin^2A....(2)

Putting B=90+A in equation(1)

cos(A+90+A)=cosAcos(90+A)-sinAsin(90+A)

-sin2A=-sinAcosA-sinAcosA

=>sin2A=2sinAcosA...(3)

Putting B=-A in equation(1)

cos(A-A)=cosAcos(-A)-sinAsin(-A)

=>cos^2A+sin^2A=1...(4)

Now finally putting B=2A in equation (1)

cos(A+2A)=cosAcos2A-sinAsin2A

inserting the relation (3)

cos3A=cosAcos2A-2sinAsinAcosA

=cosA(cos2A-2sin^2A)

Inserting relation (2)

cos3A=cosA(cos^2A-sin^2A-2sin^2A)

=cosA(cos^2A-3sin^2A)

=cosA(4cos^2A-3cos^2A-3sin^2A)

=cosA(4cos^2A-3(cos^2A+sin^2A)

=cosA(4cos^2A-3*1)" " relation(4)

cos3A=4cos^3A-3cosA

Re-arrenging and deviding bothsides by 4

:.cos^3A-3/4cosA=1/4cos3A

Proved

Jun 14, 2016

RHS is
1/4cos(3A) = 1/4cos(2A+A)
Using the given expression
=> 1/4(cos (2A)cos(A) - sin(2A)sinA)

  1. Substituting the known expression for sin (2A)=2sinA cos A
  2. Using given expression again for
    cos(2A)=cos (A+A)=cos^2A-sin^2A
  3. and writing all terms in "cosine" using sin^2A+cos^2A=1 we obtain
    1/4((2cos^2A-1)cosA - 2sin^2AcosA)

=> 1/4(2cos^3A - cosA - 2(1-cos^2A)cosA)
= 1/4(2cos^3A - cosA - 2cosA + 2cos^3A)
=>1/4( 4cos^3A - 3cosA)
=> cos^3A - 3/4 cosA=LHS