math (trig)

posted by .

Prove:
sin^2(x/2) = csc^2x - cot^2x / 2csc^2(x) + 2csc(x)cot(x)

On the right, factor the numberator as a difference of two perfect squares. In the denominator, factor out 2cscx.

You ought to prodeed rather quickly to the proof.

Respond to this Question

First Name
School Subject
Your Answer

Similar Questions

  1. Calculus

    Hello, I'm having trouble with this exercise. Can you help me?
  2. Trigonometry

    Hello all, In our math class, we are practicing the trigonometric identities (i.e., sin^2(x)+cos^2(x)=1 or cot(x)=cos(x)/sin(x). Now, we are working on proofs that two sides of an equation are equal (for example, sin(x)*csc(x)=1; sin(x)csc(x)=sin(x)/sin(x)=1; …
  3. trig

    Prove the identity sin squared 0 with line/2 = csc zero with line -cot zero with a line /2csc zero with a line
  4. trig

    verify : [sec(x) / csc(x) - cot(x)] - [sec(x) / csc(x) + cot(x)] = 2csc(x)
  5. trig

    For each expression in column I, choose the expression from column II to complete an identity: Column I Column II 1. -tanxcosx A. sin^2x/cos^2x 2. sec^2x-1 B. 1/sec^2x 3. sec x/cscx C. sin(-x) 4. 1+sin^2x D.csc^2x-cot^2x+sin^2x 5. …
  6. Math

    What is the first step. Explain please. Which expression is equivalent to cos^2x + cot^2x + sin2^x?
  7. Trig verifying identities

    I am having trouble with this problem. sec^2(pi/2-x)-1= cot ^2x I got : By cofunction identity sec(90 degrees - x) = csc x secx csc-1 = cot^2x Then split sec x and csc-1 into two fractions and multiplied both numerator and denominators …
  8. Trigonometry

    prove the ff: a.) sin(x/2) - cos(x/2) = +-sqrt(1 - sin(x)) b.) tan(x/2) + cot(x/2) = 2csc(x)
  9. TRIGO

    prove the ff: a.) sin(x/2) - cos(x/2) = +-sqrt(1 - sin(x)) b.) tan(x/2) + cot(x/2) = 2csc(x) HELP PLEASE!!
  10. Math

    Find an equation for the tangent line to the curve at (π/2 , 2). y = 4 + cot(x) - 2csc(x) I am confused how to take the derivative of this problem. When I tried to solve it I ended up with -csc^2 (x) + (2csc(x) * cot(x)). From …

More Similar Questions