Sep 9, 2014 · The quotient rule states that given functions u and v such that y = (u(x))/(v(x)), dy/dx = (u'(x)v(x) - u(x)v'(x))/(v^2(x)) By assigning u and v equal to the numerator and denominator, respectively, in the example function, we arrive at dy/dx = [(-sin x)(1+sin x) - (1+cos x)(cos x)]/(1+sin x)^2 When using the Quotient Rule, it is important to designate your functions u(x) and v(x) in such a ...

Mar 31, 2016 · Use the Pythagorean trigonometric identity, #color(blue)(sin^2x+cos^2x=1)#, to simplify "#sin^2x#" in the numerator.

Dec 26, 2016 · d/(dx) (-1/sin^2x) = (2cosx)/(sin^3x) Based on the chain rule: d/(dx) (-1/sin^x) = d/(dsinx) (-1/sin^2x) * (dsinx)/(dx)= 2/(sin^3x)cosx

Apr 14, 2018 · One cannot verify #sqrt(1+sin(x))-sqrt(1-sin(x)) =2*sin(x/2)# because it is not an identity. Explanation: I shall show that #sqrt(1+sin(x))-sqrt(1-sin(x)) =2*sin(x/2)# is not an identity by substituting the real value #x= (3pi)/4# :

Nov 5, 2016 · How do you show that #2 \sin x \cos x = \sin 2x#? is true for #(5pi)/6#? How do you prove that #sec xcot x = csc x#? How do you prove that #cos 2x(1 + tan 2x) = 1#?

Apr 13, 2016 · How do you show that #2 \sin x \cos x = \sin 2x#? is true for #(5pi)/6#? How do you prove that #sec xcot x = csc x#? How do you prove that #cos 2x(1 + tan 2x) = 1#?

In order to derive #cot(x)/(1-sin(x))#, you need to apply the quotient rule first.. The quotient rule states this:

Jan 4, 2018 · int\\ 1/sin^2(x)+cos(2x)\\ dx=1/2sin(2x)-cot(x)+C First, I will split up the integral into two parts: int\\ 1/sin^2(x)+cos(2x)\\ dx=int\\ 1/sin^2(x)\\ dx + int\\ cos(2x)\\ dx I will call the left one Integral 1 and the right one Integral 2 Integral 1 We can use the following trigonometric identity: 1/sin(theta)=csc(theta) This gives: int\\ 1/sin^2(x)\\ dx=int\\ csc^2(x)\\ dx The derivative of ...

Aug 22, 2016 · arcsin x = arcsin (1/x) The 2 solutions are: sin x = 1, and #sin (1/x) = 1# sin x = - 1 , and #sin (1/x) = - 1# Any other values of x will make the equations untrue. a. sin x = 1 --> arc x = pi/2 b. sin x = - 1 --> arc x = (3pi)/2 Answers for (0, 2pi) #pi/2, (3pi)/2# Check arc x = pi/2 --> arcsin (1) = arcsin (1/1)

Please refer to explanation below Remember : sin^2x + cos^2x = 1 2sinx cosx = sin2x Step 1: Rewrite the problem as it is 1 + sin 2x = (sin x+cosx)^2 Step 2: Pick a side you want to work on - (right hand side is more complicated) 1+ sin (2x) = (sin x+ cos x)(sin x+cosx) = sin^2x + sinx cosx +sinx cos x + cos^2x = sin^2x + 2sinx cosx + cos^2x = (sin^2x + cos^2x ) …