The cosine function has an absolute value that cannot exceed 1. Therefore the is no angle x such that cos(x) = 3. That is, there is no angle x such that x = cos^-1(3).
cos2x = 1 - sin2x = 1 - 0.64 = 0.36 So cos x = +/- 0.6 Since x is acute, cos x is +ve, so cos x = 0.6
at -90 degrees the value of cos(x) is 0.
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 this is a homework assignment, please consider trying to answer it yourself first, otherwise the value of the reinforcement of the lesson offered by the assignment will be lost on you.The deriviative d/dx cos2(x) can be evaluated using the product rule.Recall that d/dx f(x)g(x) = g(x) d/dx f(x) + f(x) d/dx g(x)In this case, both f(x) and g(x) are cos(x), and cos2x = cos(x)cos(x), so...d/dx cos(x)cos(x) = -cos(x)sin(x) - cos(x)sin(x) = -2sin(x)cos(x)
The derivative of the natural log is 1/x, therefore the derivative is 1/cos(x). However, since the value of cos(x) is submitted within the natural log we must use the chain rule. Then, we multiply 1/cos(x) by the derivative of cos(x). We get the answer: -sin(x)/cos(x) which can be simplified into -tan(x).
The cosine function has an absolute value that cannot exceed 1. Therefore the is no angle x such that cos(x) = 3. That is, there is no angle x such that x = cos^-1(3).
cos2x = 1 - sin2x = 1 - 0.64 = 0.36 So cos x = +/- 0.6 Since x is acute, cos x is +ve, so cos x = 0.6
1226
at -90 degrees the value of cos(x) is 0.
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
1 - cos x as x approaches 0. what is the cos of 0? It is 1. So as x approaches 0 cos x approaches 1. 1 - 1 = 0 So as it gets very small the solutions gets smaller.
1, 2, 257, 514:1 x 514, 2 x 257
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)
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
If this is a homework assignment, please consider trying to answer it yourself first, otherwise the value of the reinforcement of the lesson offered by the assignment will be lost on you.The deriviative d/dx cos2(x) can be evaluated using the product rule.Recall that d/dx f(x)g(x) = g(x) d/dx f(x) + f(x) d/dx g(x)In this case, both f(x) and g(x) are cos(x), and cos2x = cos(x)cos(x), so...d/dx cos(x)cos(x) = -cos(x)sin(x) - cos(x)sin(x) = -2sin(x)cos(x)