document.write( "Question 629265: Let tan(alpha) , tan(beta) be the roots of equation x^2-px+q=0, then find cos2(alpha + beta). \n" ); document.write( "
Algebra.Com's Answer #396454 by jsmallt9(3758)\"\" \"About 
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.

\n" ); document.write( "To save on typing I'm going to use A and B instead of alpha and beta.

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

\n" ); document.write( "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).

\n" ); document.write( "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:
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\n" ); document.write( "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.)
\n" ); document.write( "
\n" ); document.write( "\"tan%28A%2BB%29+=+p%2F%281-%28p%5E2+-+%28p%5E2-4q%29%29%2F4%29\"
\n" ); document.write( "\"tan%28A%2BB%29+=+p%2F%281-%284q%2F4%29%29\"
\n" ); document.write( "\"tan%28A%2BB%29+=+p%2F%281-q%29\"

\n" ); document.write( "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:
\n" ); document.write( "\"sin%28A%2BB%29+=+p%2Fsqrt%28p%5E2+%2B+%281%2Bq%29%5E2%29\"

\n" ); document.write( "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:
\n" ); document.write( "\"cos%282%28a%2BB%29%29+=+1+-+2sin%5E2%28A%2BB%29\"
\n" ); document.write( "Replacing the sin(A+B) with the expression we found earlier:
\n" ); document.write( "\"cos%282%28a%2BB%29%29+=+1+-+2%28p%2Fsqrt%28p%5E2+%2B+%281%2Bq%29%5E2%29%29%5E2\"
\n" ); document.write( "Simplifying:
\n" ); document.write( "\"cos%282%28a%2BB%29%29+=+1+-+2%28p%5E2%2F%28p%5E2+%2B+%281%2Bq%29%5E2%29%29\"
\n" ); document.write( "\"cos%282%28a%2BB%29%29+=+1+-+%282p%5E2%29%2F%28p%5E2+%2B+%281%2Bq%29%5E2%29\"
\n" ); document.write( "
\n" ); document.write( "\"cos%282%28a%2BB%29%29+=+%28-p%5E2+%2B+%281%2Bq%29%5E2%29%2F%28p%5E2+%2B+%281%2Bq%29%5E2%29\"
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