495007
Description
Flashcards by declanlarkins, updated more than 1 year ago
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Created by declanlarkins
over 10 years ago
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| Question | Answer |
| sec \(\theta\) = | \(\frac{1}{cos\theta}\) |
| cosec\(\theta\) = | \(\frac{1}{sin\theta}\) |
| cot\(\theta\) = | \(\frac{1}{tan\theta}\) |
| \(sec^2\)\(\theta\) = | 1 + \(tan^2\)\(\theta\) |
| \(cosec^2\)\(\theta\) = | 1 + \(cot^2\)\(\theta\) |
| sin2A = | 2sinAcosA |
| cos2A = | \(cos^2\)A - \(sin^2\)A |
| cos2A = | 2\(cos^2\)A - 1 |
| cos2A = | 1 - 2\(sin^2\)A |
| tan2A = | \(\frac{2tanA}{1-tan^2A}\) |
| Differentiate y=\(e^{kx}\) | \(\frac{dy}{dx}\) = k\(e^{kx}\) |
| \(\frac{1}{cos\theta}\) = | sec \(\theta\) |
| \(\frac{1}{sin\theta}\) = | cosec\(\theta\) |
| \(\frac{1}{tan\theta}\) = | cot\(\theta\) |
| 1 + \(tan^2\)\(\theta\) = | \(sec^2\)\(\theta\) |
| 1 + \(cot^2\)\(\theta\) = | \(cosec^2\)\(\theta\) |
| 2sinAcosA = | sin2A |
| \(cos^2\)A - \(sin^2\)A = | cos2A |
| 2\(cos^2\)A - 1 = | cos2A |
| 1 - 2\(sin^2\)A = | cos2A |
| \(\frac{2tanA}{1-tan^2A}\) = | tan2A |
| \(\int\)k\(e^{kx}\)\(dx\) | y=\(e^{kx}\) |
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