2019-수학-가형-홀수-18¶
문제¶
풀이¶
from sympy import *
r, theta, phi = symbols('r theta phi')
#phi = ... in terms of theta
eqn = theta+2*phi-pi/2
solve(eqn, phi)
[-theta/2 + pi/4]
#r = ... in terms of theta
AC = r
AB = 1
solve(r*cos(theta) - AB, r)
[1/cos(theta)]
BC = r*sin(theta)
CD = BC/cos(phi)
BD = CD*sin(phi)
AD = AB-BD
AE = AD
S = Rational(1,2)*AD**2*theta
ABC = Rational(1,2)*AB*BC
ABE = Rational(1,2)*AB*AE*sin(theta)
T = ABC-ABE
expr = S**2/T
expr = expr.subs({r:1/cos(theta), phi:-theta/2 + pi/4})
limit(expr, theta, 0)
\[\displaystyle \frac{1}{2}\]