No, but cos(-x) = cos(x), because the cosine function is an even function.
cos x equals cos -x because cos is an even function. An even function f is such that f(x) = f(-x).
cos(x)-cos(x)sin2(x)=[cos(x)][1-sin2(x)]cos(x)-cos(x)sin2(x)=[cos(x)][cos2(x)]cos(x)-cos(x)sin2(x)=cos3(x)
A*sin(x) + cos(x) = 1B*sin(x) - cos(x) = 1Add the two equations: A*sin(x) + B*sin(x) = 2(A+B)*sin(x) = 2sin(x) = 2/(A+B)x = arcsin{2/(A+B)}That is the main solution. There may be others: depending on the range for x.
sin(x)-cos(x) = (1)sin(x)+(-1)cos(x) so the range is sqrt((1)^2+(-1)^2)=1 and the domain is R <><><><><> The domain of sin x - cos x is [-infinity, +infinity]. The range of sin x - cos x is [-1.414, +1.414].
The derivative of cos(x) equals -sin(x); therefore, the anti-derivative of -sin(x) equals cos(x).
No, but cos(-x) = cos(x), because the cosine function is an even function.
cos x equals cos -x because cos is an even function. An even function f is such that f(x) = f(-x).
cos(x)-cos(x)sin2(x)=[cos(x)][1-sin2(x)]cos(x)-cos(x)sin2(x)=[cos(x)][cos2(x)]cos(x)-cos(x)sin2(x)=cos3(x)
A*sin(x) + cos(x) = 1B*sin(x) - cos(x) = 1Add the two equations: A*sin(x) + B*sin(x) = 2(A+B)*sin(x) = 2sin(x) = 2/(A+B)x = arcsin{2/(A+B)}That is the main solution. There may be others: depending on the range for x.
sin(x)-cos(x) = (1)sin(x)+(-1)cos(x) so the range is sqrt((1)^2+(-1)^2)=1 and the domain is R <><><><><> The domain of sin x - cos x is [-infinity, +infinity]. The range of sin x - cos x is [-1.414, +1.414].
If x = sin θ and y = cos θ then: sin² θ + cos² θ = 1 → x² + y² = 1 → x² = 1 - y²
Since the range of the cosine function is (-1,1), the function y = cos(x) assumes a minimum value of -1 for y.
sec(x)=1/cos(x), by definition of secant.
cos(60) = 0.57 x 60 x cos(60) = 7 x 30 = 210
x = 45 degrees sin(x) = cos(x) = 1/2 sqrt(2)
1. Anything divided by itself always equals 1.