Expert: Sherry Wallin - 3/6/2011

Question
Dear Sherry Wallin,

I have encountered a proving question on trigo as shown below;

Prove 2cos36 = 1+2cos72.

RHS; 1+2cos72
    =1+2cos(90-18)
    =1+2[cos(90)cos(18)+sin(90)sin(18)]
    =1+2sin(90)sin(18)
    =1+2sin(18)

    =2cos(36)  (shown)

However, I find my proofing is not concrete enough to show the above identities. I seek for your advise and the solution to the question.

Thanks

Regards
Raymond

Answer
Hi Raymond~

Are these the identities you were given? Usually identities and proofs are for arbitrary or general quantities, i.e., instead of actual numerical angles use a general form of an angle like: Prove 2cosx = 1 + 2cos2x

I don't know exactly what you mean by "my proofing isn't concrete enough to show the above identities". Using numbers is more concrete than the abstractness of an arbitrary value so I am a bit unsure what you are trying to tell me.

The reason your work works is because you are using the fact that you have a 90 degree angle which eliminates the cos 90 and makes the sin 18, the sin 18. This isn't true in general as I will show you in the following proof:

1 + 2cos 2x
= 1 + 2(2cos^2 x -1)  using the identity cos 2x = 1 - 2sin^2 x
= 1 + 4cos^2 x - 2
= -1 + 4cos^2x
 

which is not your desired conclusion because you want 2cosx. In order for this to be true all values of -1 + 4cos^2x must equal 2cosx, which means it is then an identity. So set them equal and find out where this happens:

2cosx = -1 + 4cos^2x =>
4cos^2x -2cosx -1 = 0 =>
x = (1 +- sqrt5)/4

I used the quadratic formula: x = [2+-sqrt((-2)^2-4(4)(-1))]/(2(4)) =>
x = [2+-sqrt(4+16)]/8 =>
x = [2+-2sqrt5]/8 =>
x = [2(1 +- sqrt5)/8] =>
x = (1 +- sqrt5)/4 =>
x = (1/4) +- (sqrt5)/4 =>
x = (1/4) + (sqrt5)/4, (1/4) - (sqrt5)/4 =>
x = (1+sqrt5)/4, (1-sqrt5)/4 =>
x ~= .809, -.309

So the statement is true when x = .809 and -.309 but not at any other time, and in order for this to be an identity, all values of the argument (angle) must be true, thus the two are not an identity.

Math Prof