If for $0 < x< \pi/2$, $e^{[(\sin ^{2}x+\sin ^{4}x+\sin ^{6}x+...+\infty )\log_{e}2]}$ satisfies the quadratic equation, $x^{2}-9x+8=0$, find the value of $\displaystyle \frac{\sin x-\cos x}{\sin x+\cos x}$
The infinite series sin^2x + sin^4x + ... is geometric and sums to tan^2x. Thus 2^(tan^2x) is a root of x^2 - 9x + 8 = 0; since it is greater than 1, it equals 8, giving tan^2x = 3 and x = pi/3. Therefore, (sin x - cos x)/(sin x + cos x) = 2 - sqrt(3).
The infinite geometric series inside the brackets is sine squared x plus sine fourth x and so on, which sums to sine squared x divided by (1 minus sine squared x), or tangent squared x. The expression becomes 2 raised to the power of tangent squared x, and setting it equal to the roots of the quadratic equation x squared minus 9x plus 8 equals 0, which are 1 and 8, gives two cases. For the first case, 2 raised to the power of tangent squared x equals 1 means tangent squared x is 0, which is impossible for x between 0 and pi divided by 2. For the second case, 2 raised to the power of tangent squared x equals 8 means tangent squared x is 3, so tangent of x is the square root of 3, making x equal to pi divided by 3. We must evaluate the fraction (sine of x minus cosine of x) divided by (sine of x plus cosine of x), which simplifies to (tangent of x minus 1) divided by (tangent of x plus 1). Substituting the square root of 3 gives (the square root of 3 minus 1) divided by (the square root of 3 plus 1), which rationalizes to 2 minus the square root of 3.