Problem Analysis #common_08

Recorded Python call:

import sympy as sp
θ = sp.symbols('θ')
# We have sinθ + 3cosθ = 0 => sinθ = -3cosθ
# Use identity sin^2θ + cos^2θ = 1
cosθ = sp.symbols('cosθ')
eq = sp.Eq(-3*cosθ, sp.sqrt(1 - cosθ**2))  # sinθ = -3cosθ, and sinθ = ±sqrt(1-cos^2)
# Actually, let's solve for sinθ directly
sinθ, cosθ = sp.symbols('sinθ cosθ')
eq1 = sp.Eq(sinθ + 3*cosθ, 0)
eq2 = sp.Eq(sinθ**2 + cosθ**2, 1)
solutions = sp.solve([eq1, eq2], (sinθ, cosθ), dict=True)
solutions

Recorded Python call:

import sympy as sp
sinθ, cosθ = sp.symbols('sinθ cosθ')
eq1 = sp.Eq(sinθ + 3*cosθ, 0)
eq2 = sp.Eq(sinθ**2 + cosθ**2, 1)
solutions = sp.solve([eq1, eq2], (sinθ, cosθ), dict=True)
solutions

Recorded Python call:

print('hello')

Recorded Python call:

import numpy as np
# sinθ + 3cosθ = 0 => sinθ = -3cosθ
# sin^2 + cos^2 = 1 => 9cos^2 + cos^2 = 1 => 10cos^2 = 1 => cos^2 = 1/10
cos_vals = np.array([1/np.sqrt(10), -1/np.sqrt(10)])
sin_vals = -3 * cos_vals
print("cos values:", cos_vals)
print("sin values:", sin_vals)
# Condition: cos(π - θ) = -cosθ > 0 => cosθ < 0
# So cosθ must be negative
valid = cos_vals < 0
print("valid:", valid)
print("sinθ for valid:", sin_vals[valid])

주어진 조건: $\sin\theta + 3\cos\theta = 0$이고 $\cos(\pi - \theta) > 0$.

풀이:

1. $\sin\theta + 3\cos\theta = 0$에서 $\sin\theta = -3\cos\theta$.
2. 항등식 $\sin^2\theta + \cos^2\theta = 1$에 대입하면:

따라서 $\cos\theta = \pm \frac{1}{\sqrt{10}}$, $\sin\theta = -3\cos\theta = \mp \frac{3}{\sqrt{10}}$. 3. 조건 $\cos(\pi - \theta) > 0$를 이용한다. $\cos(\pi - \theta) = -\cos\theta$이므로, $-\cos\theta > 0 \Rightarrow \cos\theta < 0$. 따라서 $\cos\theta = -\frac{1}{\sqrt{10}}$이고, $\sin\theta = -3\left(-\frac{1}{\sqrt{10}}\right) = \frac{3}{\sqrt{10}} = \frac{3\sqrt{10}}{10}$.

정답은 선택지 ①이다.