# CSC 2 FAQ

 Using the quotient rule and trigonometric identities, we can obtain the following...
 What do you mean by rounding off CSC? I on...
 The trigonometric function period for csc X/2 is 2Pi.
 csc (π/2) = 1/[sin (π/2)] = 1/1 = 1 that is really fundamental...you should know this ! ¶
 The mathematical equation csc theta + sqrt 2 =0 can be solved to find t...
 The derivative of cscU2x is -csc2xcot2x(2) ...MORE...
 -csc^2(pi/4) =.-1/sin^2(pi/4) =.-1/(sin(pi/4))^2 =.-1/(1/√2)^2 =.-1/(1/2) =.-2.
 s (t) = y = csc (t/2). let u = t/2. du/dt = 1/2. y = csc u. dy/du = - csc u cot u. dy/dt = (-1/2) csc u cot u. dy/dt = (-1/2) csc (t/2) cot (t/2).
 f(θ) = -2 csc θ. f(θ) = -2 / sinθ. f(θ) = -2 * (sinθ)^(-1). Now use the chain rule to differentiate it: f '(θ) = 2 (sinθ)^(-2) * cosθ. f '(θ) = [2 cosθ] / sin² θ. You could leave i...
 (cscx)^2*secx. =1/[(sinx)^2*cosx]. =[(sinx)^2+(cosx)^2]/ [(sinx)^2cosx]. =1/cosx + cosx/(sinx)^2. =secx + cscxcotx. Integrate: ln(secx+tanx) - cscx + C.
 d/dx csc^2 x = 2 csc x • (- csc x cot x) = -2 csc^2 x cot x
 First, let's work on the left hand side of your equation, the Csc2x This mea...
 The solution of csc (U2) x-4cotx+2 is ((partial d))/((partial d)x)(U...
 The 86th derivative of 3/2 csc x 874 is 41.
 First write tan(x/2) in terms of tan x. (You have that in your identities ...
 Using Trig identies csc (x) U2-1 is equal to cot (x) U2 or (csc(x)-1) (csc(x)+1) when fa...
 The limit of "csc (xE2) * x * sin (xE2)" as x approaches 0 is 0. Thanks for a...
 All of the solutions for csc(3x)+2=0 are x=-PI/18+2PI*n and x=-5*PI... ...MORE...
 Csc A would be equal to 2 pi. in the equation {cot(A) = 3/2&&csc(A)<0,csc(A)}.
 Example: Solve csc x + 2 = 0 for 0 <= x < 2 LaTeX Code: \\pi Choic... ...MORE...
 In the equation 2sin (x) = 3+csc (x), x is equal to 0.4387 ...MORE...
 csc θ = 2. 1/sin θ = 2. sin θ = 1/2. θ = sinˉ¹ (1/2) = π/6 + 2nπ, 5π/6 + 2nπ where n is an integer.
 Integral(csc^2(x))dx= Integral(dx/sin^2(x)) divide num and denom by cos^2(x) to get. Integral(1/cos^2(x)/(sin^2(x)/cos^2x))… (sec^2(x)/tan^2(x))dx. let tan(x)=u sec^2(x)dx=du so. y...
 so sin ( 4 Θ ) = - 1 / 2 ---> 4 Θ = 7 π / 6 + 2 n π or - π / 6 + 2 n π---->. Θ = 7π / 28 + nπ / 2 or - π / 24 + nπ / 2...you finish.
 Csc x is defined as. 1/(sin x). So, 1/(sin^2 x) must be equal to csc^2 x. What don't you understand ? QED.
 arccos(1 / 2) = pi / 3. csc(pi / 3) = 2 sqrt(3) / 3.
 I think in the description you mean cscx, not cotx. But other than that you're completely right. think of it this way.-csc²x = -(cscx)² = -(1/sinx)² = -(1²/(sinx)²) = -1/sin²x.
 You have x = csc^-1(√2). so csc(x) = √2. sin(x) = 1/√2 = √2/2. So x = π/4.
 y = csc x/2. csc is the reciprocal of sin, so where sin (x/2) = 0, csc (x/2) would have vertical asymptotes. At x = 0, 2π, sin x/2 = 0, so there are vertical asymptotes at 0 and 2π...
 ∫(4 - csc²(x)) dx = 4x + cot(x) + C