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Find the range of f(theta) = sqrt(8 sin²(theta) + 4 cos²(theta) - 8 sin(theta) cos(theta)).

1. [sqrt5 - 1, sqrt5 + 1]
2. [0, sqrt5 + 1]
3. [sqrt6 - sqrt20, sqrt6 + sqrt20]
4. none of these

Correct answer: [sqrt5 - 1, sqrt5 + 1]

Solution

8 sin² + 4 cos² - 8 sin cos = 4 + 4 sin² - 4 sin(2theta) = 4 + 2(1 - cos2theta) - 4 sin2theta = 6 - 2cos2theta - 4sin2theta. The amplitude of (-2cos2theta - 4sin2theta) is sqrt(4 + 16) = sqrt20 = 2sqrt5. So the inside ranges over [6 - 2sqrt5, 6 + 2sqrt5] = [(sqrt5 - 1)², (sqrt5 + 1)²]. Taking square roots gives range [sqrt5 - 1, sqrt5 + 1].

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