sec2 = 1/cos(2) = -2.403 (radians)
sec4(Θ) - sec2(Θ) = tan4(Θ) + tan2(Θ)Factor each side: sec2(Θ) [sec2(Θ) - 1] = tan2(Θ) [tan2(Θ) + 1]Use the identitiy: 1 + tan2(Θ) = sec2(Θ)We can also write it as: tan2(Θ) = sec2(Θ) - 1Substitute the first form of the identity in the right side: sec2(Θ) [sec2(Θ) - 1] = tan2(Θ) sec2(Θ)Substitute the second form of the identity in the left side: sec2(Θ) tan2(Θ) = tan2(Θ) sec2(Θ)Is that close enough for you ?Is it worth a trust point ?
It is sec2.
1 kilometer = 1,000 meters 1 hour = 3,600 seconds 1 meter/sec2 = (1 meter/sec2) x (1 kilometer/1,000 meters) x (3,6002 sec2/hr2) = 12,960 km/hr2 1 km/hr2 = 7.716 x 10-5 meter/sec2
The derivative of a function is the function's tangent function.So, d(y)/dx = d(2tan(2x))/dx = 2*d(tan(2x))/dx = 2*d(tan(u))/du*du/dx, where u=2x, = 2*sec2(2x)*d(2x)/dx = 4*sec2(2x)*d(x)/dx = 4*sec2(2x)Now just make a plot for y = 4*sec2(2x) and you got your tangent function.
H= -1/2gt2+vt+s Where H is the ending height g is the rate of gravity (32 ft/sec2 or 9.8 m/sec2) t is the time v is the initial velocity and s is the starting height.
sec4(Θ) - sec2(Θ) = tan4(Θ) + tan2(Θ)Factor each side: sec2(Θ) [sec2(Θ) - 1] = tan2(Θ) [tan2(Θ) + 1]Use the identitiy: 1 + tan2(Θ) = sec2(Θ)We can also write it as: tan2(Θ) = sec2(Θ) - 1Substitute the first form of the identity in the right side: sec2(Θ) [sec2(Θ) - 1] = tan2(Θ) sec2(Θ)Substitute the second form of the identity in the left side: sec2(Θ) tan2(Θ) = tan2(Θ) sec2(Θ)Is that close enough for you ?Is it worth a trust point ?
derivative of sec2(x)=2tan(x)sec2(x)
It is sec2.
km/hr2 x 7.71604938 x 10^-8) = km/sec2
9.6 m/sec2.
1 kilometer = 1,000 meters 1 hour = 3,600 seconds 1 meter/sec2 = (1 meter/sec2) x (1 kilometer/1,000 meters) x (3,6002 sec2/hr2) = 12,960 km/hr2 1 km/hr2 = 7.716 x 10-5 meter/sec2
One fourth of a gram, or 2.45 m/sec2 .
5.052 feet/sec2
There is a trigonometric identity that states that sec2(x) - tan2(x) = 1, for every x. By rearranging this formula we can find that sec2(x) - 1 = tan2(x).
The formula for calculating force (F) in newtons is F mass (m) x acceleration (a). In this case, with a mass of 5 kg and acceleration of 2 m/sec2, the force can be calculated as F 5 kg x 2 m/sec2 10 newtons.
The derivative of a function is the function's tangent function.So, d(y)/dx = d(2tan(2x))/dx = 2*d(tan(2x))/dx = 2*d(tan(u))/du*du/dx, where u=2x, = 2*sec2(2x)*d(2x)/dx = 4*sec2(2x)*d(x)/dx = 4*sec2(2x)Now just make a plot for y = 4*sec2(2x) and you got your tangent function.
G is the 7th letter of the modern English alphabet In most science texts "g" is used to designate the acceleration due to gravity at Earthh's surface (9.8 m/sec2 or 32 ft/sec2)