0
To analyze the function ( F(x) = x \sin x + \cos x ) over the intervals ( [0, 2\pi] ) and ( [\pi, 2\pi] ), we can evaluate its behavior by determining critical points and endpoints within those intervals. The first derivative, ( F'(x) = \sin x + x \cos x - \sin x = x \cos x ), helps identify where the function is increasing or decreasing. In the interval ( [0, 2\pi] ), ( F(x) ) generally increases due to positive contributions from ( x \cos x ) in the first quadrant, while in ( [\pi, 2\pi] ), the behavior is influenced by the negative cosine term leading to a decrease. Ultimately, analyzing the endpoints and critical points will provide specific values and insights into the function's behavior on those intervals.
What else can I help you with?
What is the solution to sec plus tan equals cos over 1 plus sin?
sec + tan = cos /(1 + sin) sec and tan are defined so cos is non-zero. 1/cos + sin/cos = cos/(1 + sin) (1 + sin)/cos = cos/(1 + sin) cross-multiplying, (1 + sin)2 = cos2 (1 + sin)2 = 1 - sin2 1 + 2sin + sin2 = 1 - sin2 2sin2 + 2sin = 0 sin2 + sin = 0 sin(sin + 1) = 0 so sin = 0 or sin = -1 But sin = -1 implies that cos = 0 and cos is non-zero. Therefore sin = 0 or the solutions are k*pi radians where k is an integer.
How do you solve sinx divided by 1 plus cosx plus cosx divided by sinx?
sin x/(1+cos x) + cos x / sin x Multiply by sin x (1+cos x) =[(sin^2 x + cos x(1+cos x) ] / sin x (1+cos x) = [(sin^2 x + cos x + cos^2 x) ] / sin x (1+cos x) sin^2 x + cos^2 x = 1 = (1+cos x) / sin x (1+cos x) = 1/sin x
How do you prove this trigonometric relationship sin3A equals 3sinA cos 2 A - sin 3 A?
sin(3A) = sin(2A + A) = sin(2A)*cos(A) + cos(2A)*sin(A)= sin(A+A)*cos(A) + cos(A+A)*sin(A) = 2*sin(A)*cos(A)*cos(A) + {cos^2(A) - sin^2(A)}*sin(A) = 2*sin(A)*cos^2(A) + sin(a)*cos^2(A) - sin^3(A) = 3*sin(A)*cos^2(A) - sin^3(A)
Sin30 cos90 sin 90 cos30?
To simplify the expression sin(30°) cos(90°) sin(90°) cos(30°), we first evaluate the trigonometric functions at the given angles. sin(30°) = 1/2, cos(90°) = 0, sin(90°) = 1, and cos(30°) = √3/2. Substituting these values into the expression, we get (1/2) * 0 * 1 * (√3/2) = 0. Therefore, the final result of sin(30°) cos(90°) sin(90°) cos(30°) is 0.
What is the derivative of 6 sin x plus 2 cos x?
F(x) = 6 sin(x) + 2 cos(x)F'(x) = 6 cos(x) - 2 sin(x)
Verify that sin minus cos plus 1 divided by sin plus cos subtract 1 equals sin plus 1 divided by cos?
[sin - cos + 1]/[sin + cos - 1] = [sin + 1]/cosiff [sin - cos + 1]*cos = [sin + 1]*[sin + cos - 1]iff sin*cos - cos^2 + cos = sin^2 + sin*cos - sin + sin + cos - 1iff -cos^2 = sin^2 - 11 = sin^2 + cos^2, which is true,
What is the solution to sec plus tan equals cos over 1 plus sin?
sec + tan = cos /(1 + sin) sec and tan are defined so cos is non-zero. 1/cos + sin/cos = cos/(1 + sin) (1 + sin)/cos = cos/(1 + sin) cross-multiplying, (1 + sin)2 = cos2 (1 + sin)2 = 1 - sin2 1 + 2sin + sin2 = 1 - sin2 2sin2 + 2sin = 0 sin2 + sin = 0 sin(sin + 1) = 0 so sin = 0 or sin = -1 But sin = -1 implies that cos = 0 and cos is non-zero. Therefore sin = 0 or the solutions are k*pi radians where k is an integer.
Find the values of θ in the range 0 to 2π for cos θ plus cos 3θ equals sin θ plus sin 3θ?
The angle can be 0, pi/2, pi, 3*pi/2 or 2*pi radians.
How do you factor sin squared times x plus cos2x -cosx equals 0?
2
How do you solve sinx divided by 1 plus cosx plus cosx divided by sinx?
sin x/(1+cos x) + cos x / sin x Multiply by sin x (1+cos x) =[(sin^2 x + cos x(1+cos x) ] / sin x (1+cos x) = [(sin^2 x + cos x + cos^2 x) ] / sin x (1+cos x) sin^2 x + cos^2 x = 1 = (1+cos x) / sin x (1+cos x) = 1/sin x
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
Sin x plus cos x?
sin(x) + cos(x) = sqrt(2) · sin(45°+x)
How do you prove this trigonometric relationship sin3A equals 3sinA cos 2 A - sin 3 A?
sin(3A) = sin(2A + A) = sin(2A)*cos(A) + cos(2A)*sin(A)= sin(A+A)*cos(A) + cos(A+A)*sin(A) = 2*sin(A)*cos(A)*cos(A) + {cos^2(A) - sin^2(A)}*sin(A) = 2*sin(A)*cos^2(A) + sin(a)*cos^2(A) - sin^3(A) = 3*sin(A)*cos^2(A) - sin^3(A)
Sin30 cos90 sin 90 cos30?
To simplify the expression sin(30°) cos(90°) sin(90°) cos(30°), we first evaluate the trigonometric functions at the given angles. sin(30°) = 1/2, cos(90°) = 0, sin(90°) = 1, and cos(30°) = √3/2. Substituting these values into the expression, we get (1/2) * 0 * 1 * (√3/2) = 0. Therefore, the final result of sin(30°) cos(90°) sin(90°) cos(30°) is 0.
What is sin of 3 pi over 2?
Sin(3pi/2) = Sin(2pi - pi/2) Double angle Trig. Identity. Hence Sin(2pi)Cos(pi/2) - Cos(2pi) Sin(pi/2) Sin(2pi) = 0 Cos(pi/2) = 0 Cos(2pi) = 1 Sin(pi/2) = 1 Substituting 0 x 0 - 1 x 1 = 0 - 1 = -1 The answer!!!!!
What is the derivative of 6 sin x plus 2 cos x?
F(x) = 6 sin(x) + 2 cos(x)F'(x) = 6 cos(x) - 2 sin(x)
What is cosx-2sinxcosx equals 0?
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.
