cos x - 1 = 0 cos(x) = 1 x = 0 +/- k*pi radians where k = 1,2,3,...
It would be 1 over square root 5.
Secant x= 1/cosx So if cos x=1 ,we know that x=0 degrees ( or radians), so secant x is 1/cos (0)=1/1=1
22
y = cos ( x ). z = cos ( x + 180 degrees ).
Sin[x] = Cos[x] + (1/3)
1/ Tan = 1/ (Sin/Cos) = Cos/Sin = Cot (Cotangent)
If x = sin θ and y = cos θ then: sin² θ + cos² θ = 1 → x² + y² = 1 → x² = 1 - y²
1. Anything divided by itself always equals 1.
No, but cos(-x) = cos(x), because the cosine function is an even function.
cos x - 1 = 0 cos(x) = 1 x = 0 +/- k*pi radians where k = 1,2,3,...
-1
f(x)= sin(1/x) and g(x)=1/sin(x) [u(v)]' = u'(v) * v', where u and v are functions So f'(x) = sin'(1/x) * (1/x)' = cos(x) * (-1/x2) = -cos(x)/x2 g'(x) = (1/x)' applied to sin(x) * (sin(x))' = -1/(sin2(x)) * cos(x) = -cos(x)/(sin2(x))
It would be 1 over square root 5.
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)
1
No, (sinx)^2 + (cosx)^2=1 is though