SOLUTION: Trig problem has me stumped. Express as a cotangent function the following: {y=sqrt((1+(sqrt(2)/2)(cos(xπ/3)+sin(xπ/3)))/(1-(sqrt(2)/2)(cos(xπ/3)+sin(xπ/3))))} Almost too

Algebra ->  Trigonometry-basics -> SOLUTION: Trig problem has me stumped. Express as a cotangent function the following: {y=sqrt((1+(sqrt(2)/2)(cos(xπ/3)+sin(xπ/3)))/(1-(sqrt(2)/2)(cos(xπ/3)+sin(xπ/3))))} Almost too       Log On


   



Question 1195395: Trig problem has me stumped. Express as a cotangent function the following:
{y=sqrt((1+(sqrt(2)/2)(cos(xπ/3)+sin(xπ/3)))/(1-(sqrt(2)/2)(cos(xπ/3)+sin(xπ/3))))}
Almost too complicated to put in a single string. Please let me know if I can send you an image of the problem. Thank you.

Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!

y%22%22=%22%22

let sqrt%282%29%2F2%22%22=%22%22a

let cos%28%28x%2Api%29%2F3%29%2Bsin%28%28x%2Api%29%2F3%29%22%22=%22%22b

Substituting:

y%22%22=%22%22sqrt%28%281%2Bab%29%2F%281-ab%29%29

Under the radical multiply top and bottom by the conjugate
of the denominator, as if you were rationalizing the denominator:

y%22%22=%22%22sqrt%28%28%281%2Bab%29%281%2Bab%29%29%2F%28%281-ab%29%281%2Bab%29%29%29

y%22%22=%22%22sqrt%28%28%281%2Bab%29%5E2%29%2F%281-%28ab%29%5E2%29%29

The numerator is a perfect square, so we take the square root of
the numerator and denominator:

y%22%22=%22%22%281%2Bab%5E%22%22%29%2Fsqrt%281-%28ab%29%5E2%29%29

Write the right side as the sum of two fractions:

y%22%22=%22%22%281%5E%22%22%29%2Fsqrt%281-%28ab%29%5E2%29%29%22%22%2B%22%22%28ab%5E%22%22%29%2Fsqrt%281-%28ab%29%5E2%29%29

Next we draw a right triangle with an angle θ, 1 as the hypotenuse, ab
as the adjacent side and the radical as the opposite side:



So now the equation is 

y%22%22=%22%22csc%28theta%29%22%22%2B%22%22cot%28theta%29 where cos%28theta%29%22%22=%22%22ab

Since you want y as a function of cot(θ), we use an identity for csc(θ).

1 + cot2(θ) = csc2(θ), solve for csc(θ)

    csc%28theta%29%22%22=%22%22sqrt%281%2Bcot%5E2%28theta%29%29

And we substitute back for ab:

The final equation is 

y%22%22=%22%22sqrt%281%2Bcot%5E2%28theta%29%29%22%22%2B%22%22cot%28theta%29 where  cos%28theta%29%22%22=%22%22%28sqrt%282%29%2F2%29%28cos%28%28x%2Api%29%2F3%29%2Bsin%28%28x%2Api%29%2F3%29%5E%22%22%29


Edwin