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What is the derivative of -2sin x?
The derivative of sin(x) is cos(x). Coefficients act like constants and always remain in derivatives. So, the derivative is -2cos(x).
How do you prove sin x tan x equals cos x?
You can't. tan x = sin x/cos x So sin x tan x = sin x (sin x/cos x) = sin^2 x/cos x.
Sin x Tan x equals Sin x?
No. Tan(x)=Sin(x)/Cos(x) Sin(x)Tan(x)=Sin2(x)/Cos(x) Cos(x)Tan(x)=Sin(x)
Solution for tan x plus cot x divided by sec x csc x?
(tan x + cot x)/sec x . csc x The key to solve this question is to turn tan x, cot x, sec x, csc x into the simpler form. Remember that tan x = sin x / cos x, cot x = 1/tan x, sec x = 1/cos x, csc x = 1/sin x The solution is: [(sin x / cos x)+(cos x / sin x)] / (1/cos x . 1/sin x) [(sin x . sin x + cos x . cos x) / (sin x . cos x)] (1/sin x cos x) [(sin x . sin x + cos x . cos x) / (sin x . cos x)] (sin x . cos x) then sin x. sin x + cos x . cos x sin2x+cos2x =1 The answer is 1.
How do you solve the following identity sec x - cos x equals sin x tan x?
sec x - cos x = (sin x)(tan x) 1/cos x - cos x = Cofunction Identity, sec x = 1/cos x. (1-cos^2 x)/cos x = Subtract the fractions. (sin^2 x)/cos x = Pythagorean Identity, 1-cos^2 x = sin^2 x. sin x (sin x)/(cos x) = Factor out sin x. (sin x)(tan x) = (sin x)(tan x) Cofunction Identity, (sin x)/(cos x) = tan x.
What are the coefficients of the equation 2x-5y-23?
2 and 5 are coefficients of x and y respectively.
What is the expansion of angle sin?
The expansion as in Taylor series for sin(x) is: sin(x) = x - (x^3)/3! + (x^5)/5! - (x^7)/7! + ...
What is the derivative of -2sin x?
The derivative of sin(x) is cos(x). Coefficients act like constants and always remain in derivatives. So, the derivative is -2cos(x).
When are y equals sin x and y equals cos x times cos x equal?
y=sin x y=cos x cos x sin x = cos^2 x sin x = 1-sin^2 x sin x -1 + sin^2 x = 0 sin^2 x + sin x -1 = 0 Let y=sin x y^2+y-1 = 0 This equation is of form ay^2+by+c=0 a = 1 b = 1 c = -1 y=[-b+/-sqrt(b^2-4ac)]/2a] y=[-1 +/-sqrt(1^2-4(1)(-1)]/(2)(1) discriminant is b^2-4ac =5 y=[-1 +√(5)] / 2 y=[-1 -√(5)] / 2 sin x = [-1 +√(5)] / 2 x = sin^-1 [-1 +√(5)] / 2] = 0.6662394 radians x = sin^-1 [-1 -√(5)] / 2] = sin^-1 (-1.618) -- has no solution When x = 0.6662394 radians, sin x and cos x times cos x are equal.
How do you solve sin parenthesis cos to the negative 1 parenthesis 2 over 5 closed parenthesis and another closed parenthesis?
Let y = sin(cos-1(2/5)) Suppose x = cos-1(2/5): that is, cos(x) = 2/5 then sin2(x) = 1 - cos2(x) = 1 - 4/25 = 21/25 so that sin(x) = sqrt(21)/5 which gives x = sin-1[sqrt(21)/5] Then y = sin(cos-1(2/5)) = sin(x) : since x = cos-1(2/5) =sin{sin-1[sqrt(21)/5]} = sqrt(21)/5 There will be other solutions that are cyclically related to this one but no range has been given for the solutions.
Can polynomials have radicals?
no only the coefficients can. like rad 5 but not x or y
What is multiplicand mean in terms of math?
The quantity to be multiplied by the multiplier is called multiplicand. Say sin(x) is multiplied by 5. Then sin(x) is the multiplicand. 5 is the multiplier. If sin(x) is the multiplier, then 5 would become the multiplicand.
How do you solve 2 sin2-5 sin plus 2 equals 0?
2*sin^2(x) - 5*sin(x) + 2 = 0 is a quadratic equation in sin(x).therefore,{2*sin(x) - 1}*{sin(x) - 2)} = 0=> sin(x) = 1/2 or sin(x) = 2The second solution is rejected since sin(x) cannot exceed 1.The principal solution is x = arcsin(1/2) = pi/6 radians. Additional or alternative solutions will depend on the domain for x - which has not been given.
What is x if the sin of x equals square root of 5 divided by 2?
2.5
Find the derivative y equals 10 to the sin3x power?
Y=10^sin(x) The derivative is: (log(5)+log(2))*cos(x)*2^sin(x)*5^sin(x) Use the chain rule, product rule, and power rules combined with sin(x) rule.
What is the amplitude for y equals 5 sin x?
5
Solution for tan x is equal to cos x?
Tan(x) = Sin(x) / Cos(x) Hence Sin(x) / Cos(x) = Cos(x) Sin(x) = Cos^(2)[x] Sin(x) = 1 - Sin^(2)[x] Sin^(2)[x] + Sin(x) - 1 = 0 It is now in Quadratic form to solve for Sin(x) Sin(x) = { -1 +/-sqrt[1^(2) - 4(1)(-1)]} / 2(1) Sin(x) = { -1 +/-sqrt[5[} / 2 Sin(x) = {-1 +/-2.236067978... ] / 2 Sin(x) = -3.236067978...] / 2 Sin(x) = -1.61803.... ( This is unresolved as Sine values can only range from '1' to '-1') & Sin(x) = 1.236067978... / 2 Sin(x) = 0.618033989... x = Sin^(-1) [ 0.618033989...] x = 38.17270765.... degrees.
