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Home » the allele for black noses in wolves is dominant - Sourci

the allele for black noses in wolves is dominant - Sourci

📅 May 1, 2026 👤 scraface

May 01, 2026

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📌 Solution: The expression $ \cos x + i\sin x $ corresponds to $ e^{ix} $ on the unit circle in the complex plane. Therefore, $ \cos x + i\sin x + 1 = e^{ix} + 1 $. The modulus is $ |e^{ix} + 1| $. Using the identity $ |e^{ix} + 1| = \sqrt{(\cos x + 1)^2 + \sin^2 x} $, we simplify:

📌 |e^{ix} + 1| = \sqrt{(\cos x + 1)^2 + \sin^2 x} = \sqrt{\cos^2 x + 2\cos x + 1 + \sin^2 x} = \sqrt{2 + 2\cos x}

📌 The minimum value occurs when $ \cos x $ is minimized, i.e., $ \cos x = -1 $:

📌 \sqrt{2 + 2(-1)} = \sqrt{0} = 0

📌 Thus, the minimum value is $ \boxed{0} $.

📌 Question: Find the range of the function $ f(x) = \frac{1}{\sin^2 x + 4\cos^2 x} $ as $ x $ ranges over all real numbers such that $ \sin^2 x + \cos^2 x = 1 $.

📌 Solution: Since $ \sin^2 x + \cos^2 x = 1 $, we write $ f(x) = \frac{1}{\sin^2 x + 4\cos^2 x} = \frac{1}{1 + 3\cos^2 x} $. Let $ y = \cos^2 x $, so $ 0 \leq y \leq 1 $. Then $ f(x) = \frac{1}{1 + 3y} $. As $ y $ varies from 0 to 1, $ 1 + 3y $ varies from 1 to 4, so $ f(x) $ varies from $ \frac{1}{4} $ to $ 1 $. Hence, the range of $ f(x) $ is $ \left[\frac{1}{4}, 1\right] $. The final answer is $ \boxed{\left[\frac{1}{4}, 1\right]} $.

📌 Question: Find all angles $ z \in [0^\circ, 360^\circ] $ such that $ \tan z + \cot z = 2 $.

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