SOLUTION: Let tan(alpha) , tan(beta) be the roots of equation x^2-px+q=0, then find cos2(alpha + beta).

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: Let tan(alpha) , tan(beta) be the roots of equation x^2-px+q=0, then find cos2(alpha + beta).       Log On


   

Question 629265: Let tan(alpha) , tan(beta) be the roots of equation x^2-px+q=0, then find cos2(alpha + beta).
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
There may be be a much easier way to do this. But since you haven't gotten a response in over a day, I'd thought I'd give it a try.

To save on typing I'm going to use A and B instead of alpha and beta.

Using the Quadratic Formula on x%5E2-px%2Bq=0 we get:
x+=+%28-%28-p%29+%2B-+sqrt%28%28-p%29%5E2-4%281%29%28q%29%29%29%2F2%281%29
Simplifying...
x+=+%28-%28-p%29+%2B-+sqrt%28p%5E2-4%281%29%28q%29%29%29%2F2%281%29
x+=+%28-%28-p%29+%2B-+sqrt%28p%5E2-4q%29%29%2F2%281%29
x+=+%28p+%2B-+sqrt%28p%5E2-4q%29%29%2F2
which is short for
x+=+%28p+%2B+sqrt%28p%5E2-4q%29%29%2F2 or x+=+%28p+-+sqrt%28p%5E2-4q%29%29%2F2

It doesn't matter which one we say is tan(A) and which one is tan(B). So I'll make the first root tan(A) and the second one tan(B).

Next we use the formula for tan%28A%2BB%29+=+%28tan%28A%29+%2B+tan%28B%29%29%2F%281-tan%28A%29%2Atan%28B%29%29. Inserting our roots into this formula we get:

Simplifying... (Note: to multiply the two numerators in the denominator, I am going to use the %28a%2Bb%29%28a-b%29+=+a%5E2-b%5E2 pattern.)

tan%28A%2BB%29+=+p%2F%281-%28p%5E2+-+%28p%5E2-4q%29%29%2F4%29
tan%28A%2BB%29+=+p%2F%281-%284q%2F4%29%29
tan%28A%2BB%29+=+p%2F%281-q%29

From this we are going to find an expression for sin(A+B). tan is opposite/adjacent. So draw a right triangle, pick one of the acute angles and make the opposite side "p" and the adjacent side "1+q". Then use the Pythagorean Theorem to find the hypotenuse. You should get sqrt%28p%5E2+%2B+%281%2Bq%29%5E2%29. Since sin is opposite/hypotenuse:
sin%28A%2BB%29+=+p%2Fsqrt%28p%5E2+%2B+%281%2Bq%29%5E2%29

Next we will use the cos%282x%29+=+1+-+2sin%5E2%28x%29 variation of the cos(2x) formulas to find our answer. Replacing the x's with (A+B)'s:
cos%282%28a%2BB%29%29+=+1+-+2sin%5E2%28A%2BB%29
Replacing the sin(A+B) with the expression we found earlier:
cos%282%28a%2BB%29%29+=+1+-+2%28p%2Fsqrt%28p%5E2+%2B+%281%2Bq%29%5E2%29%29%5E2
Simplifying:
cos%282%28a%2BB%29%29+=+1+-+2%28p%5E2%2F%28p%5E2+%2B+%281%2Bq%29%5E2%29%29
cos%282%28a%2BB%29%29+=+1+-+%282p%5E2%29%2F%28p%5E2+%2B+%281%2Bq%29%5E2%29

cos%282%28a%2BB%29%29+=+%28-p%5E2+%2B+%281%2Bq%29%5E2%29%2F%28p%5E2+%2B+%281%2Bq%29%5E2%29