You could just pull out the half: it will be (1/2) cos squared x.
Yes. Except where sin x = 0, because then you would be dividing by zero so the quotient is undefined.
One solution. (cos x)2 - 2cos x = 3 Factor: (cos x - 3)(cos x + 1)= 0 cos x = {-1, 3} Solve: For cos x = -1, x = 180 deg No solution for cos x = 3
The deriviative of sin2 x + cos2 x is 2 cos x - 2 sin x
No, (sinx)^2 + (cosx)^2=1 is though
2 x cosine squared x -1 which also equals cos (2x)
No. Cos squared x is not the same as cos x squared. Cos squared x means cos (x) times cos (x) Cos x squared means cos (x squared)
Cos^2 x = 1 - sin^2 x
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
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.
Sin squared, cos squared...you removed the x in the equation.
22
You can look up "trigonometric identities" in Wikipedia.Cos(2x), among other things, is equal to (cos x)^2 - (sin x)^2 If you meant cos squared x, or (cos x)^2, that is equal to (1 + cos(2x))/2
Sin squared is equal to 1 - cos squared.
You could just pull out the half: it will be (1/2) cos squared x.
Yes. Except where sin x = 0, because then you would be dividing by zero so the quotient is undefined.
One solution. (cos x)2 - 2cos x = 3 Factor: (cos x - 3)(cos x + 1)= 0 cos x = {-1, 3} Solve: For cos x = -1, x = 180 deg No solution for cos x = 3