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Note: Conversion is based on the latest values and formulas.
How do you find the general solutions for cos^2(x) - Socratic 19 Oct 2015 · Substitute (5) in (1) to find #cos(x)# Once you have the final answer do not forget that this cyclic. So it repeats every turn of a circle which is #2pi# radians.
How do you find the indefinite integral of #int 2^(sinx)cosx#? 24 Dec 2016 · #color(white)(int 2^(sin x) cos x " "dx)=1/(ln 2) * 2^u+C# And since #u = sin x# , we substitute back: #color(white)(int 2^(sin x) cos x " "dx)=1/(ln 2) * 2^sin x+C#
How do you differentiate y=2 csc x + 5 cos x? | Socratic 20 Mar 2018 · Calculus Differentiating Trigonometric Functions Derivative Rules for y=cos(x) and y=tan(x) 1 Answer
Is f(x)=cos(-x) increasing or decreasing at x=0? - Socratic 6 Apr 2018 · f(x)=cos(-x) is neither increasing nor decreasing at x=0 The first derivative provides the slope at each point of our function. A positive slope means the function is increasing and that a negative slope means that the function is decreasing. Knowing this, we can find out if cos(-x) is increasing or decreasing at color(red)(x=0) by evaluating its first derivative at that point. Before …
How do you verify Cos(x+(pi/6))- cos(x-(pi/6))=1? - Socratic 24 Sep 2016 · How do you find all the solutions for #2 \sin^2 \frac{x}{4}-3 \cos \frac{x}{4} = 0# over the
How do you find the derivative of #f(x) = (sin x)(cos x)#? - Socratic 21 Jul 2016 · #(d f(x))/(d x) (sin x)( cos x)=?# #d/(d x) sin x=cos x# #d/(d x)cos x=-sin x# #y=a*b" ; "y^'=a^'*b+b^'*a# #(d f(x))/(d x) (sin x)( cos x)=cos x*cos x-sin x*sin x#
What is the domain and range of #y = cos|x|#? - Socratic 13 Jul 2016 · The domain is (-infty, infty) and the range is [-1,1]. This is a fun problem because we are presented with a modified version of the cos(x) function, but as we will see, it is, in fact, not in any way different from the standard version. cos|x| is the cos(x) function with the absolute value of x input into it. What this means is that if x >= 0 then it is replaced with x. If x < 0 then it is ...
What is the derivative of #sin((pi/2) - x)#? - Socratic 30 Aug 2016 · -sinx sin(pi/2-x) = cos x, so d/dx(sin(pi/2-x) ) = d/dx(cosx) = -sinx If, for some reason, you don't want to simplify the expression, use d/dx sin(u) = cos u * (du ...
How do you find the general solutions for cos 2x + cos x - 2 =0? 12 Aug 2015 · How do you find the general solutions for #cos 2x + cos x - 2 =0#? Trigonometry Trigonometric Identities and Equations Solving Trigonometric Equations
Provet that? : cos(-x) -= cosx | Socratic 25 Oct 2016 · This is a proof We use the sum of angles formula: cos(A+-B) -= cosAcosB+-sinAsinB Put A=0 and B=-x to get cos(0-x) -= cos0cosx-sin0sinx And we know that sin0=0 and cos0=1 so: cos(-x) -= (1)cosx - 0 :. cos(-x) -= cosx QED