sec(x)=1/cos(x), by definition of secant.
Cos(90 - x) = sin(x) so cos2(90 - x) = sin2(x)
cos x - 1 = 0 cos(x) = 1 x = 0 +/- k*pi radians where k = 1,2,3,...
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That means you must take the derivative of the derivative. In this case, you must use the product rule. y = 6x sin x y'= 6[x (sin x)' + (x)' sin x] = 6[x cos x + sin x] y'' = 6[x (cos x)' + (x)' cos x + cos x] = 6[x (-sin x) + cos x + cos x] = 6[-x sin x + 2 cos x]
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)
If x = sin θ and y = cos θ then: sin² θ + cos² θ = 1 → x² + y² = 1 → x² = 1 - y²
sec(x)=1/cos(x), by definition of secant.
7.5 x 8 = 60
x = 45 degrees sin(x) = cos(x) = 1/2 sqrt(2)
1. Anything divided by itself always equals 1.
Prove that tan(x)sin(x) = sec(x)-cos(x) tan(x)sin(x) = [sin(x) / cos (x)] sin(x) = sin2(x) / cos(x) = [1-cos2(x)] / cos(x) = 1/cos(x) - cos2(x)/ cos(x) = sec(x)-cos(x) Q.E.D
1
Cos(90 - x) = sin(x) so cos2(90 - x) = sin2(x)