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Home » Subtract the first from the second: - Blask

Subtract the first from the second: - Blask

📅 March 9, 2026 👤 scraface

Mar 09, 2026

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📌 \sec \theta = \frac{1}{\cos \theta}, \quad \csc \theta = \frac{1}{\sin \theta}.

📌 \sec \theta + \csc \theta = \frac{1}{\cos \theta} + \frac{1}{\sin \theta} = \frac{\sin \theta + \cos \theta}{\sin \theta \cos \theta}.

📌 The square is:

📌 Using the identity \((\sin \theta + \cos \theta)^2 = 1 + \sin 2\theta\), and \(\sin \theta \cos \theta = \frac{1}{2} \sin 2\theta\), we rewrite it as:

📌 Let \(y = \sin 2\theta\), where \(y \in (0, 1]\). Define:

📌 f(y) = \frac{4(1 + y)}{y^2}.

📌 To find the maximum of \(f(y)\) on \( (0,1] \), take the derivative:

📌 f'(y) = 4 \cdot \frac{(1)(y^2) - (1 + y)(2y)}{y^4} = 4 \cdot \frac{y^2 - 2y(1 + y)}{y^4} = 4 \cdot \frac{y^2 - 2y - 2y^2}{y^4} = 4 \cdot \frac{-y^2 - 2y}{y^4} = -4 \cdot \

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Question: A science policy analyst is modeling the growth of renewable...
g(x) = \left(\sqrt{2} \sin x + \frac{1}{\sqrt{2} \sin x}\right)^2
Find the minimum value of $ g(x) $ for $ x $ such that $ \sin x > 0 $.
We expand the expression:
g(x) = \left(\sqrt{2} \sin x + \frac{1}{\sqrt{2} \sin x}\right)^2 = 2...
Let $ y = \sin^2 x $, where $ y > 0 $. Then:
g(x) = 2y + 2 + \frac{1}{2y}
We minimize $ 2y + \frac{1}{2y} + 2 $. Let:
h(y) = 2y + \frac{1}{2y}
Using the AM-GM inequality:

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