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What else can I help you with?
Sine cubed plus cosine cubed divided by sine plus cosine?
[sin(x)^3 + cos(x)^3] / [sin(x) + cos(x)]= [(sin(x) + cos(x))(sin(x)^2 - sin(x)cos(x) + cos(x)^2)] / [sin(x) + cos(x)]***Now you can cancel a "sin(x) + cos(x)" from the top and bottom of the fraction. This makes the bottom of the fraction equal to 1. I am just going to write the next step without a 1 on the bottom of the fraction (x/1=x).So now you just have:= (sin(x)^2 - sin(x)cos(x) + cos(x)^2) *I'm going to move some terms around now. ~Not doing any computation in this step.= (sin(x)^2 + cos(x)^2 - sin(x)cos(x)) *Now we know that cos(x)^2 + sin(x)^2 = 1.= 1 - sin(x)cos(x)
2 sin x - 3 equals 0?
2 sin(x) - 3 = 0 2 sin(x) = 3 sin(x) = 1.5 No solution. The maximum value of the sine function is 1.0 .
What is the derivative of sin x plus 2?
d/dx [sin(x) + 2] = cos(x)
What is x minus 3 divided by x squared minus x minus 6?
-3
What is the factor of 10 x squared minus X minus 2?
10x^2 - x - 2 = (5x + 2)(2x - 1)
How do you show that 2 sin squared x minus 1 divided by sin x minus cos x equals sin x plus cos x?
(2 sin^2 x - 1)/(sin x - cos x) = sin x + cos x (sin^2 x + sin^2 x - 1)/(sin x - cos x) =? sin x + cos x [sin^2 x - (1 - sin^2 x)]/(sin x - cos x) =? sin x + cos x (sin^2 x - cos^2 x)/(sin x - cos x) =? sin x + cos x [(sin x - cos x)(sin x + cos x)]/(sin x - cos x) =? sin x + cos x sin x + cos x = sin x + cos x
2 sin squared x plus sin x minus 1 is equal to 0 solve for x?
2 sin2(x) + sin(x) - 1 = 0(2 sin + 1) (sin - 1) = 0Either 2 sin(x) + 1 = 02sin(x) = -1sin(x) = -0.5x = 210°, 330°or sin(x) - 1 = 0sin(x) = 1x = 90°
What is the limit of sin multiplied by x minus 1 over x squared plus 2 as x approaches infinity?
As X approaches infinity it approaches close as you like to 0. so, sin(-1/2)
What is the answer to cos square x divide by 1 minus sin x?
cos2 x /(1 - sin x)= (1 - sin2 x )/(1 - sin x)= (1 + sin x)(1 - sin x)/(1 - sin x)= 1 + sin x
What is sin squared x minus cos squared x?
Sin squared, cos squared...you removed the x in the equation.
How do you solve 2 sin squared x minus sinx minus 1 is equal to 0?
2 sin(x)2 - sin(x) - 1 = 0 Let Y=sin(x) then the equation is 2*Y2 - Y - 1 =0 Delta = (-1 * -1) - 4 * 2 * -1 = 9 Y = (1 + sqrt(9)) / 4 or Y = (1 - sqrt(9)) / 4 Y = 1 or Y = -1/2 Then x = Arcsin(Y) and (in radians) x = Arcsin(1) = Pi/2 +2*k*Pi or x=Arcsin(-1/2) = -Pi/6 + 2*k*Pi where k is an integer
How do you simplify this expression... cot2x over csc2x-cscx... please note that cot2x and csc2x are really just raised to the second power?
I assume the expression is cot^2 x / ( csc^2 x - csc x) express it in terms of sin x and cos x: =(cos^2 x / sin^2 x) / (1/sin^2 x - 1/sin x) =(cos^2 x / sin^2 x) / [(1 - sin x)/sin^2 x] =cos^2 x / (1 - sin x) = (1 - sin^2 x) / (1 - sin x) = (1 + sin x)(1 - sin x) / (1 - sin x) = 1 + sin x
Verify sin squared x minus sin squared y equals sin x plus y times sin x-y?
To verify sin2x - sin2y = sin x + y sin (x-y) [if I've read your equation correctly] is impossible (for all x and y). A counter example can be easily found whereby the values of the two halves of the (supposed) equality are different. Let x = π and y = π/2. Then: Left hand side: sin2x - sin2y = sin2π - sin2 π/2= 0 - 12 = -1 Right hand side: sin x + y sin (x-y) = sin π + π/2 sin (π - π/2) = 0 - π/2 sin (π/2) = -π/2 Thus sin2x - sin2y = -1 ≠ -π/2 = sin x + y sin (x-y) when x = π and y = π/2, so the equality cannot be verified -
What is the first deriviative of sinsquaredx?
If f(x) = sin^2(x) then f'(x) = 2 sin(x) d/dx(sin(x)) = 2 sin(x) cos(x) = sin(2x)
How do you factor sin squared times x plus cos2x -cosx equals 0?
2
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.
Find the derivative of y x2 sin x 2xcos x - 2sin x?
y = (x^2)(sin x)(2x)(cos x) - 2sin xy' = [[(x^2)(sin x)][(2x)(cos x)]]' - (2sin x)'y' = [[(x^2)(sin x)]'[(2x)(cos x)] + [(2x)(cos x)]'[(x^2)(sin x)]]- (2sin x)'y' = [[(x^2)'(sin x) + (sin x)'(x^2)][(2x)(cos x)] + [(2x)'(cos x) + (cos x)'(2x)][(x^2)(sin x)] ] - 2(cos x)y' = [[(2x)(sin x )+ (cos x)(x^2)][(2x)(cos x)] + [2cos x - (sin x)(2x)][(x^2)(sin x)]] - 2(cos x)y' = (4x^2)(sin x cos x) + (2x^3)(cos x)^2 + (2x^2)(sin x cos x) - (2x^3)(sin x)^2 - 2cos xy' = (6x^2)(sin x cos x) + (2x^3)(cos x)^2 - (2x^3)(sin x)^2 - 2cos x (if you want, you can stop here, or you can continue)y' = (3x^2)(2sin x cos x) + (2x^3)[(cos x)^2 - (sin x)^2] - 2cos xy' = (3x^2)(sin 2x) + (2x^3)(cos 2x) - 2 cos xy' = (2x^3)(cos 2x) + (3x^2)(sin 2x) - 2 cos x
