0
Add your answer:
Earn +20 pts
How do you get the csc theta given tan theta in quadrant 1?
If tan(theta) = x then sin(theta) = x/(sqrt(x2 + 1) so that csc(theta) = [(sqrt(x2 + 1)]/x = sqrt(1 + 1/x2)
What is the anti-derivative of secant squared?
negative cotangent -- dcot(x)/dx=-csc^2(x)
What is the range of F(x)dividing x?
The range of F(x)dividing x depends on the domain of x and on the function F.
How do you simplify csc theta tan theta?
With all due respect, you don't really want to know howto solve it.You just want the solution.csc(Θ) = 1/sin(Θ)tan(Θ) = sin(Θ)/cos(Θ)csc(Θ) x tan(Θ) = 1/sin(Θ) x sin(Θ)/cos(Θ) = 1/cos(Θ) = sec(Θ)
Definition of the 6 trigonometric functions?
The six main trigonometric functions are sin(x)=opposite/hypotenuse cos(x)=adjacent/hypotenuse tan(x)=opposite/adjacent csc(x)=hypotenuse/opposite cot(x)=adjacent/opposite sec(x)=hypotenuse/adjacent Where hypotenuse, opposite, and adjacent correspond to the three sides of a right triangle and x corresponds to an angle in that right triangle.
What is the derivative of csc x?
The derivative of csc(x) is -cot(x)csc(x).
What is tan x csc x?
tan(x)*csc(x) = sec(x)
What is the derivative of cscx?
d/dx csc(x) = - csc(x) tan(x)
How are the graphs of sec x and csc x related?
They are co-functions meaning that 90 - sec x = csc x.
Verify that Cos theta cot theta plus sin theta equals csc theta?
It's easiest to show all of the work (explanations/identities), and x represents theta. cosxcotx + sinx = cscx cosx times cosx/sinx + sinx = csc x (Quotient Identity) cosx2 /sinx + sinx = csc x (multiplied) 1-sinx2/sinx + sinx = csc x (Pythagorean Identity) 1/sinx - sinx2/sinx + sinx = csc x (seperate fraction) 1/sinx -sinx + sinx = csc x (canceled) 1/sinx = csc x (cancelled) csc x =csc x (Reciprocal Identity)
What is the derivative of 3tanx-4cscx?
7
How do you simplify csc theta cot theta?
There are 6 basic trig functions.sin(x) = 1/csc(x)cos(x) = 1/sec(x)tan(x) = sin(x)/cos(x) or 1/cot(x)csc(x) = 1/sin(x)sec(x) = 1/cos(x)cot(x) = cos(x)/sin(x) or 1/tan(x)---- In your problem csc(x)*cot(x) we can simplify csc(x).csc(x) = 1/sin(x)Similarly, cot(x) = cos(x)/sin(x).csc(x)*cot(x) = (1/sin[x])*(cos[x]/sin[x])= cos(x)/sin2(x) = cos(x) * 1/sin2(x)Either of the above answers should work.In general, try converting your trig functions into sine and cosine to make things simpler.
What is sec squared x times csc x divided by sec squared x plus csc squared x?
Ah, secant, annoying as always. Why don't we use its definition as 1/cos x and csc as 1/sin x? We will do that Also, please write down the equation, there is at least TWO different equations you are talking about. x^n means x to the power of n 1/(sin x) ^2 is csc squared x, it's actually csc x all squared 1/(cos x) ^2 in the same manner.
Does csc -120 equal -csc 120?
-240
Csc squared divided by cot equals csc x sec. can someone make them equal?
cot(x)=1/tan(x)=1/(sin(x)/cos(x))=cos(x)/sin(x) csc(x)=1/sin(x) sec(x)=1/cos(x) Therefore, (csc(x))2/cot(x)=(1/(sin(x))2)/cot(x)=(1/(sin(x))2)/(cos(x)/sin(x))=(1/(sin(x))2)(sin(x)/cos(x))=(1/sin(x))*(1/cos(x))=csc(x)*sec(x)
Csc x tan x?
If you want to simplify that, it usually helps to express all the trigonometric functions in terms of sines and cosines.
What is the anti derivative of cscxcotx?
∫cscxcotx*dx∫csc(u)cot(u)*du= -csc(u)+C, where C is the constant of integrationbecause d/dx(csc(u))=-[csc(u)cot(u)],so d/dx(-csc(u))=csc(u)cot(u).∫cscxcotx*dxLet:u=xdu/dx=1du=dx∫cscucotu*du= -csc(u)+CPlug in x for u.∫cscxcotx*dx= -csc(x)+C
