Sin squared, cos squared...you removed the x in the equation.
(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
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.
The derivative of cos(x) equals -sin(x); therefore, the anti-derivative of -sin(x) equals cos(x).
Note that an angle should always be specified - for example, 1 - cos square x. Due to the Pythagorean formula, this can be simplified as sin square x. Note that sin square x is a shortcut of (sin x) squared.
Cos^2 x = 1 - sin^2 x
Sin squared, cos squared...you removed the x in the equation.
sin2 x = (1/2)(1 - cos 2x) cos2 x = (1/2)(1 + cos 2x) Multiplying both you get (1/4) (1 - cos2 2x) Which is equal to (1/4) (1 - (1/2) (1 + cos 4x) = (1/8) (2 - 1 - cos 4x) = (1/8) (1 - cos 4x) Or If it is the trigonomic function, sin squared x and cosine squared x is equal to one
22
Yes. Except where sin x = 0, because then you would be dividing by zero so the quotient is undefined.
The deriviative of sin2 x + cos2 x is 2 cos x - 2 sin 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
(1+cosx)(1-cosx)= 1 +cosx - cosx -cos^2x (where ^2 means squared) = 1-cos^2x = sin^2x (sin squared x)
(1+cosx)(1-cosx)= 1 +cosx - cosx -cos^2x (where ^2 means squared) = 1-cos^2x = sin^2x (sin squared x)
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.