SOLUTION: please help prove 1+cosx/sinx=cot(x/2) i tried starting with: 1+cosx/sinx=cos/sin(x/2) =+/-square root 1+cosx/2/sin then i got confused

Algebra ->  Trigonometry-basics -> SOLUTION: please help prove 1+cosx/sinx=cot(x/2) i tried starting with: 1+cosx/sinx=cos/sin(x/2) =+/-square root 1+cosx/2/sin then i got confused      Log On


   

Question 232801: please help prove 1+cosx/sinx=cot(x/2)
i tried starting with:
1+cosx/sinx=cos/sin(x/2)
=+/-square root 1+cosx/2/sin
then i got confused
Answer by jsmallt9(3758) About Me  (Show Source):
You can put this solution on YOUR website!
%281%2Bcos%28x%29%29%2Fsin%28x%29+=+cot%28x%2F2%29
  • The easiest way to prove this is to use one of the identities for tan(x/2). If you are allowed to use tan(x/2) then you can use any of the following:
    • tan%28x%2F2%29+=+sin%28x%29%2F%281%2Bcos%28x%29%29
    • tan%28x%2F2%29+=+%281-cos%28x%29%29%2Fsin%28x%29
    • tan%28x%2F2%29+=+0+%2B-+sqrt%28%281-cos%28x%29%29%2F%281%2Bcos%28x%29%29%29 (Sorry about the extra 0 but Algebra.com's formula software doesn't handle "+-" without a number or variable in fron of it.)
    If we are llowed to use one of these, we can use the first one in combination with cot%28x%2F2%29+=+1%2Ftan%28x%2F2%29 to solve this very quickly:
    %281%2Bcos%28x%29%29%2Fsin%28x%29+=+1%2Ftan%28x%2F2%29
    Using the first tan(x/2) above:
  • If we're not allowed to use tan(x/2), you have a made a good first step because it is often a good idea in problems like this to rewrite expressions in terms of sin and cos.
    %281%2Bcos%28x%29%29%2Fsin%28x%29+=+cos%28x%2F2%29%2Fsin%28x%2F2%29
    Replacing cos(x/2) and sin(x/2), another good idea:

    Now where do we go from here? First we can use the property of square roots sqrt%28a%29%2Fsqrt%28b%29+=+sqrt%28a%2Fb%29 to rewrite the right side as a single square root:

    By combining the two square roots into one we can now simplify and manipulate the compound fraction within. To start the denominators of "2" cancel:

    Now what? The key here is to recognize one of the fundamental identities, %28sin%28x%29%29%5E2+=+1+-+%28cos%28x%29%29%5E2 as a difference of squares. Using the difference of squares factoring pattern, a%5E2+=+b%5E2+=+%28a%2Bb%29%28a-b%29, on the right side of the identity we get:
    %28sin%28x%29%29%5E2+=+%281+%2B+cos%28x%29%29%281+-+cos%28x%29%29
    Looking at this and looking at the square root in your equation, we can see the two factors in the identity in the numerator and denominator. So let's see what happens if we multiply the numerator and denominator by (1+cos(x)):

    Multiplying the fractions we get:

    Substituting %28sin%28x%29%29%5E2 for %281+%2B+cos%28x%29%29%281+-+cos%28x%29%29 in the denominator:

    Now the numerator and denominator are both perfect squares. So we can remove the square root:
    %281%2Bcos%28x%29%29%2Fsin%28x%29+=+%281%2Bcos%28x%29%29%2Fsin%28x%29
    And we're done.