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Why does cos x equal cos -x?
cos x equals cos -x because cos is an even function. An even function f is such that f(x) = f(-x).
How do you solve triangle EFG given that f equals 15 g equals 18 and f equals 29?
I assume you mean EFG are the interior angles you need to find and efg are the sides( must be sides as they do not add to 180). Use law of cosines. Need to change to your letters; twice. Then for third angle subtract from 180 degrees I assume you mean e = 29 ( you need to be more careful in mathematics ) DEGREE MODE! f^2 = g^2 + e^2 - 2ge cos(F) 15^2 = 18^2 + 29^2 - 2(18)(29) cos(F) 225 = 1165 - 1044(cos F) -940 = -1044(cos F) 0.90038 = cos F arcos(0.90038) = F F = 26 degrees g^2 = e^2 + f^2 - 2ef cos(G) 18^2 = 29^2 + 15^2 - 2(29)(15) cos(G) 324 = 1066 - -870(cos G) -742 = -870(cos G) 0.85287 = cos G arcos(0.85287) = G G = 31 degrees Now just subtract these answers from 180 degrees for E 180 - 26 - 31 E = 123 degrees F = 26 degrees G = 31 degrees
If fx equals cossinx2 then f prime equals?
f(x)=cos(sin(x2)) [u(v)]' = u'(v) * v' so f'(x) = cos'(sinx(x2)) * sin'(x2) * (x2)' f'(x) = -sin(sin(x2)) * cos(x2) * 2x = -2x sin(sin(x2)) cos(x2)
What is the period of the function f(x) cos 2x?
The function ( f(x) \cos 2x ) has a period determined by the cosine component. The cosine function ( \cos 2x ) has a period of ( \frac{2\pi}{2} = \pi ). Therefore, regardless of the form of ( f(x) ), the overall function ( f(x) \cos 2x ) will also have a period of ( \pi ), assuming ( f(x) ) does not introduce any additional periodicity.
What is the derivative of 6 sin x plus 2 cos x?
F(x) = 6 sin(x) + 2 cos(x)F'(x) = 6 cos(x) - 2 sin(x)
If f(x) cos(2x).cos(4x).cos(6x).cos(8x).cos(10x) then find the limit of 1 - f(x)35(sinx)2 as x tends to 0 (zero).?
I'm sorry the question is not correctly displayed. If f(x) = cos(2x).cos(4x).cos(6x).cos(8x).cos(10x) then, find the limit of {1 - [f(x)]^3}/[5(sinx)^2] as x tends to 0 (zero).
Why does cos x equal cos -x?
cos x equals cos -x because cos is an even function. An even function f is such that f(x) = f(-x).
Why is cos an even function?
An even function is one where f(x) = f(-x) For cosine, cos(x) = cos(-x), thus cosine is an even function.
How do you solve triangle EFG given that f equals 15 g equals 18 and f equals 29?
I assume you mean EFG are the interior angles you need to find and efg are the sides( must be sides as they do not add to 180). Use law of cosines. Need to change to your letters; twice. Then for third angle subtract from 180 degrees I assume you mean e = 29 ( you need to be more careful in mathematics ) DEGREE MODE! f^2 = g^2 + e^2 - 2ge cos(F) 15^2 = 18^2 + 29^2 - 2(18)(29) cos(F) 225 = 1165 - 1044(cos F) -940 = -1044(cos F) 0.90038 = cos F arcos(0.90038) = F F = 26 degrees g^2 = e^2 + f^2 - 2ef cos(G) 18^2 = 29^2 + 15^2 - 2(29)(15) cos(G) 324 = 1066 - -870(cos G) -742 = -870(cos G) 0.85287 = cos G arcos(0.85287) = G G = 31 degrees Now just subtract these answers from 180 degrees for E 180 - 26 - 31 E = 123 degrees F = 26 degrees G = 31 degrees
If fx equals cossinx2 then f prime equals?
f(x)=cos(sin(x2)) [u(v)]' = u'(v) * v' so f'(x) = cos'(sinx(x2)) * sin'(x2) * (x2)' f'(x) = -sin(sin(x2)) * cos(x2) * 2x = -2x sin(sin(x2)) cos(x2)
What is the period of the function f(x) cos 2x?
The function ( f(x) \cos 2x ) has a period determined by the cosine component. The cosine function ( \cos 2x ) has a period of ( \frac{2\pi}{2} = \pi ). Therefore, regardless of the form of ( f(x) ), the overall function ( f(x) \cos 2x ) will also have a period of ( \pi ), assuming ( f(x) ) does not introduce any additional periodicity.
What is the derivative of 6 sin x plus 2 cos x?
F(x) = 6 sin(x) + 2 cos(x)F'(x) = 6 cos(x) - 2 sin(x)
What does cos omak mean?
cos= vagina omak= mother's cos omak = Mother's vagina Basically its, f**k your mother in arabic.
How do you differentiate cos x all squared?
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)
How can you find the theta if the given is just the coefficient friction and the mass?
If by theta you mean the angle at the base of slope on which is the body laying, and you want to calculate minimal theta for which the blocks starts to slide: Let's first calculate: weight: Q = mg force normal to the slope: N = Q cos theta = mg cos theta force tangent to the slope: F = Q sin theta = mg sin theta force of friction: T = fN = fmg cos theta, where f is coefficient of friction The body will start to move downwards, when T = F, or: fmg cos theta = mg sin theta which after simplyfying becomes: f cos theta = sin theta, f = sin theta / cos theta f = tan theta Therefore, theta = arc tan f As you see, the angle only depends on friction coefficient f. (If that's not a problem you asked to be solved, edit your question please to precisely state what needs to be calculated)
What is dydx of sincostanx?
If y = sin(cos(tan(x))) Using the chain rule: (f(g(x)))' = f'(g(x)).g'(x) Then dy/dx = cos(cos(tan(x))).-sin(tan(x)).sec2(x) = -cos(cos(tan(x))).sin(tan(x)).sec2(x) Unfortunately I don't think this can be simplified much more. ( sec = 1/cos )
If the resultant of two vectors each of magnitide f is twice of magnitude F then angle bw?
If the resultant of two vectors, each of magnitude ( f ), is twice the magnitude ( F ), then the angle ( \theta ) between the two vectors can be determined using the formula for the resultant of two vectors: [ R = \sqrt{f^2 + f^2 + 2f^2 \cos \theta} ] Given that ( R = 2F ), we set ( R = 2f ) (assuming ( F = f )). This leads to the equation ( 4f^2 = 2f^2(1 + \cos \theta) ). Solving for ( \theta ), we find that ( \cos \theta = 0 ), which means ( \theta = 90^\circ ). Thus, the angle between the vectors is ( 90^\circ ).
