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1. ### If cosθ = -5/13 and 0 <= θ <= 2π, what are possible values of θ?

Homework Statement If cosθ = -5/13 and 0 <= θ <= 2π, what are possible values of θ? Homework Equations The Attempt at a Solution θ is in quadrant 2 or 3 in quadrant 2: related acute = cos^{-1}(5/13) = 1.18 θ = π - 1.18 = 1.96 radians in quadrant 3: related acute angle = 1.18 θ = π +...
2. ### Finding the radian value of this angle which passes through a point

Homework Statement The terminal arm of an angle in standard position passes through (-7,8). Find the radian value of the angle in the interval [0,2∏], to the nearest hundredth. Homework Equations sinθ = \frac{y}{r} cosθ = \frac{x}{r} tanθ = \frac{y}{x} The Attempt at a Solution The terminal...
3. ### Finding explicit formula of this recursion formula

is that calculus? :\$
4. ### Finding explicit formula of this recursion formula

Homework Statement Write an explicit formula for the sequence determined by the following recursion formula. t_{1}= 0; t_{n} = t_{n-1} + \frac{2}{n(n+1)} The Attempt at a Solution t_{1} = 0 t_{2} = t_{1} + \frac{2}{2(2+1)} t_{2} = \frac{1}{3} t_{3} = t_{2} + \frac{2}{3(3+1)} t_{3} =...
5. ### Proving This Trigonometric Identity

=\frac{1+cosx}{1-cosx} =\frac{1+cosx}{1-cosx} × \frac{1+cosx}{1+cosx} = \frac{1+2cosx + cos^2x}{1-cos^2x} =\frac{1+2cosx+cos^2x}{sin^2x} thanks guys
6. ### Proving This Trigonometric Identity

1. Prove: \frac{cscx +cotx}{cscx-cotx} = \frac{1+2cosx+cos^2x}{sin^2x} Homework Equations tanx = \frac{sinx}{cosx} cotx = \frac{cosx}{sinx} cscx secx cotx sin^2x + cos^2x = 1 The Attempt at a Solution Left side: =\frac{cscx +cotx}{cscx-cotx} =\frac{1/sinx + cosx / sinx}{1/sinx -...
7. ### Domain and Range of this function

Ah so when x = ±2, it is undefined. So... domain: {x∈R | x ≠ ±2} What do I do for range? when i plug it in i get undefined?
8. ### Domain and Range of this function

Homework Statement What is the domain and range of the function y = \frac{2}{x^2-4} The Attempt at a Solution I'm lost, and I noticed it seems like a difference of squares so what I did is y = \frac{2}{(x-2)(x+2)}. So the x value of vertex is 0. And i subbed this into the equation...