When does sin x equal 4 cos x?

When does cotangent equal -1?

cot[x]= -1 cot[x] = cos[x] / sin[x] cos[x] / sin[x] = -1 cos[x] = -sin[x] |cos[x]| = |sin[x]| at every multiple of Pi/4 + Pi/2. However, the signs disagree at 3Pi/4 + nPi, where n is an integer.

When does cos x equal -sin x?

The derivative of cos(x) equals -sin(x); therefore, the anti-derivative of -sin(x) equals cos(x).

Why does the derivative of sin x equal - cos x?

It isn't. The derivate of sin x = cos x.It isn't. The derivate of sin x = cos x.It isn't. The derivate of sin x = cos x.It isn't. The derivate of sin x = cos x.

How do you prove that 2 sin 3x divided by sin x plus 2 cos 3x divided by cos x equals 8 cos 2x?

You need to know the trigonometric formulae for sin and cos of compound angles. sin(x+y) = sin(x)*cos(y)+cos(x)*sin(y) and cos(x+y) = cos(x)*cos(y) - sin(x)*sin(y) Using these, y = x implies that sin(2x) = sin(x+x) = 2*sin(x)cos(x) and cos(2x) = cos(x+x) = cos^2(x) - sin^2(x) Next, the triple angle formulae are: sin(3x) = sin(2x + x) = 3*sin(x) - 4*sin^3(x) and cos(3x) = 4*cos^3(x) - 3*cos(x) Then the left hand side = 2*[3*sin(x) - 4*sin^3(x)]/sin(x) + 2*[4*cos^3(x) - 3*cos(x)]/cos(x) = 6 - 8*sin^2(x) + 8cos^2(x) - 6 = 8*[cos^2(x) - sin^2(x)] = 8*cos(2x) = right hand side.

Simplify sin x plus sin x cotx equals cscx?

First convert everything to sines and cosines:sin x + sin x cos x / sin x = 1 / sin xsin x + cos x = 1 / sin xMultiplying by sin x:sin2x + sin x cos x = 1Using the identity sin2 + cos2x = 1:sin2x + sin x cos x = sin2x + cos2xsin x cos x = cos2xDividing by cos x:sin x = cos xThe solution is therefore x = pi / 4 radians, or x = 5 pi / 4 radians.The division by cos x assumed that cos x was not equal to zero; this possibility must be explored in the original equation. When cos x = 0, sin x = 1 or -1, and the angle x = pi/2 or -pi/2. It seems both of these are solutions, too.First convert everything to sines and cosines:sin x + sin x cos x / sin x = 1 / sin xsin x + cos x = 1 / sin xMultiplying by sin x:sin2x + sin x cos x = 1Using the identity sin2 + cos2x = 1:sin2x + sin x cos x = sin2x + cos2xsin x cos x = cos2xDividing by cos x:sin x = cos xThe solution is therefore x = pi / 4 radians, or x = 5 pi / 4 radians.The division by cos x assumed that cos x was not equal to zero; this possibility must be explored in the original equation. When cos x = 0, sin x = 1 or -1, and the angle x = pi/2 or -pi/2. It seems both of these are solutions, too.First convert everything to sines and cosines:sin x + sin x cos x / sin x = 1 / sin xsin x + cos x = 1 / sin xMultiplying by sin x:sin2x + sin x cos x = 1Using the identity sin2 + cos2x = 1:sin2x + sin x cos x = sin2x + cos2xsin x cos x = cos2xDividing by cos x:sin x = cos xThe solution is therefore x = pi / 4 radians, or x = 5 pi / 4 radians.The division by cos x assumed that cos x was not equal to zero; this possibility must be explored in the original equation. When cos x = 0, sin x = 1 or -1, and the angle x = pi/2 or -pi/2. It seems both of these are solutions, too.First convert everything to sines and cosines:sin x + sin x cos x / sin x = 1 / sin xsin x + cos x = 1 / sin xMultiplying by sin x:sin2x + sin x cos x = 1Using the identity sin2 + cos2x = 1:sin2x + sin x cos x = sin2x + cos2xsin x cos x = cos2xDividing by cos x:sin x = cos xThe solution is therefore x = pi / 4 radians, or x = 5 pi / 4 radians.The division by cos x assumed that cos x was not equal to zero; this possibility must be explored in the original equation. When cos x = 0, sin x = 1 or -1, and the angle x = pi/2 or -pi/2. It seems both of these are solutions, too.

Why is 2 sin theta cos theta equal to sin 2theta?

because sin(2x) = 2sin(x)cos(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

Solution for tan x is equal to cos x?

if tan x = cos x then sin x / cos x = cos x => sin x = cos x cos x => sin x = cos2 x => sin x = 1 - sin2x => sin2x + sin x - 1 = 0 Using the quadratic formula => 1. sin x = 0.61803398874989484820458683436564 => x = sin-1 (0.61803398874989484820458683436564) or => 2. sin x = -1.6180339887498948482045868343656 => x = sin-1 (-1.6180339887498948482045868343656)

What is Cos squared x equal to?

Cos^2 x = 1 - sin^2 x

What is the exact solution to cosx equals sin2x?

Cos(x) = Sin(2x) Using angle-addition, we have Sin(a+b) = Sin(a)Cos(b) + Sin(b)Cos(a). From that, we see Sin(2x) = Sin(x)Cos(x)+Sin(x)Cos(x) = 2Sin(x)Cos(x) Cos(x) = 2Sin(x)Cos(x) If Cos(x) = 0, then the two sides are equal. This occurs at x= Pi/2 + nPi, where n is an integer and Pi is approximately 3.14. If Cos(x) doesn't equal 0, then we can divide it out. Then, 1 = 2 Sin(x) , or 1/2 = Sin(x) This occurs when x = Pi/6 or 5Pi/6, plus or minus any multiples of 2 Pi.

What does cosx divided by 1-sinx equal?

cos x / (1-sin x) = cos x (1 + sin x) / (1 - sin x) (1 + sin x) = cos x (1 + sin x) / (1 - sin2x) = cos x (1 + sin x) / cos2 x = (1 + sin x) / cos x = sec x + tan xcos x / (1-sin x) = cos x (1 + sin x) / (1 - sin x) (1 + sin x) = cos x (1 + sin x) / (1 - sin2x) = cos x (1 + sin x) / cos2 x = (1 + sin x) / cos x = sec x + tan xcos x / (1-sin x) = cos x (1 + sin x) / (1 - sin x) (1 + sin x) = cos x (1 + sin x) / (1 - sin2x) = cos x (1 + sin x) / cos2 x = (1 + sin x) / cos x = sec x + tan xcos x / (1-sin x) = cos x (1 + sin x) / (1 - sin x) (1 + sin x) = cos x (1 + sin x) / (1 - sin2x) = cos x (1 + sin x) / cos2 x = (1 + sin x) / cos x = sec x + tan x

What is Cos squared x - 1 equal to?

Cos2(x-1) is equal to: 1/2 * (1 + Cos(2 - 2x)) (Cos(x) * Cos(1) - Sin(x) * Sin(1))2 1/4 * (2 + e2i - 2ix + e2ix - 2i) where e is the natural log and i is the imaginary unit.