sec(x)=1/cos(x), by definition of secant.
cos(60) = 0.57 x 60 x cos(60) = 7 x 30 = 210
cos x - 2 sin x cos x = 0 -> cos x (1 - 2 sin x) = 0 => cos x = 0 or 1 - 2 sin x = 0 cos x = 0: x = π/2 + kπ 1 - 2 sin x = 0: sin x = 1/2 -> x = π/6 + 2kπ or 5/6π + 2kπ Thus x = π/2 + kπ; x = π/6 + 2kπ; x = 5/6π + 2kπ solve the original equation.
That means you must take the derivative of the derivative. In this case, you must use the product rule. y = 6x sin x y'= 6[x (sin x)' + (x)' sin x] = 6[x cos x + sin x] y'' = 6[x (cos x)' + (x)' cos x + cos x] = 6[x (-sin x) + cos x + cos x] = 6[-x sin x + 2 cos x]
y = 2(x) - (pi/3) + (sqrt(3)/2)
1
x = 45 degrees sin(x) = cos(x) = 1/2 sqrt(2)
No, (sinx)^2 + (cosx)^2=1 is though
1
y equals x-4 plus 2 is the same as y = x-2. You just translate the graph of y=x, 2 units to the right, OR 2 down.
tan^2(x) Proof: cos^2(x)+sin^2(x)=1 (Modified Pythagorean theorem) sin^2(x)=1-cos^2(x) (Property of subtraction) cos^2(x)-1/cos^2(x)=? sin^2(x)/cos^2(x)=? (Property of substitution) sin(x)/cos(x) * sin(x)/cos(x) = tan(x) * tan(x) (Definition of tanget) = tan^2(x)
cos(2x) = 1 - 2(sin(x)^2), so sin(x)^2 = 1/2 - 1/2*cos(2x).
(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
sqrt(2)*cos(x + pi/4) [with x in radians], or sqrt(2)*cos(x + 90°) [with x in degrees]
2
cos2(x) - cos(x) = 2 Let y = cos(x) then y2 - y = 2 or y2 - y - 2 = 0 factorising, (y - 2)(y + 1) = 0 that is y = 2 or y = -1 Substitutng back, this would require cos(x) = 2 or cos(x) = -1 But cos(x) cannot be 2 so cos(x) = -1 Then x = cos-1(-1) => x = pi radians.
Yes, it is. the basic identity is for a double angle relation: cos 2x = 2 cosx cos x -1 since sec x =1/cos x if we multiply both sides by sec x we get cos2xsec x = 2cosxcos x/cos x -1/cos x = 2cos x - sec x