cos x equals cos -x because cos is an even function. An even function f is such that f(x) = f(-x).
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
Because the slope of the curve of sin(x) is cos(x). Or, equivalently, the limit of sin(x) over x tends to cos(x) as x tends to zero.
As a first step, I would convert everything to sines and cosines. sin x cot x = sec x - cos x thus becomes: (sin x) (cos x / sin x) = (1 / cos x) - cos x Simplifying: cos x = 1 / cos x - cos x It doesn't look as though they are equal. In fact, if you do the calculations for some specific angle, e.g. 30 degrees, you see that they are not.
Rewrite sec x as 1/cos x. Then, sec x sin x = (1/cos x)(sin x) = sin x/cos x. By definition, this is equal to tan x.
The values depends on what x is. This function sqrt(2)*cos(x - pi/4) [with x in radians], is equivalent to cos(x) + sin(x). or sqrt(2)*cos(x - 90°) [with x in degrees]
Cos^2 x = 1 - sin^2 x
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
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
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.
2 x cosine squared x -1 which also equals cos (2x)
Start on the left-hand side. cos(x) + tan(x)sin(x) Put tan(x) in terms of sin(x) and cos(x). cos(x) + [sin(x)/cos(x)]sin(x) Multiply. cos(x) + sin2(x)/cos(x) Make the denominators equal. cos2(x)/cos(x) + sin2(x)/cos(x) Add. [cos2(x) + sin2(x)]/cos(x) Use the Pythagorean Theorem to simplify. 1/cos(x) Since 1/cos(x) is the same as sec(x)- the right-hand side- the proof is complete.
Cos(2x) = cos^2(x) - sin^2(x) = 2 cos^2(x) - 1 = 1 - 2 sin^2(x).Source: ChaCha.com
sin2x / (1-cos x) = (1-cos2x) / (1-cos x) = (1-cos x)(1+cos x) / (1-cos x) = (1+cos x) sin2x=1-cos2x as sin2x+cos2x=1 1-cos2x = (1-cos x)(1+cos x) as a2-b2=(a-b)(a+b)
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
Because the slope of the curve of sin(x) is cos(x). Or, equivalently, the limit of sin(x) over x tends to cos(x) as x tends to zero.
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.
tan x