Question

Calculate \( \tan(75^{\circ}) \) using the difference formula.

Ask by Hills Garza. in Mozambique
Jan 22,2025

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Tutor-Verified Answer

Answer

\( \tan(75^{\circ}) = 2 + \sqrt{3} \)

Solution

Para calcular \( \tan(75^{\circ}) \) utilizando a fórmula da soma dos ângulos, podemos seguir os passos abaixo: ### Passo 1: Escolher Ângulos Conhecidos Escolhemos dois ângulos cuja soma seja \(75^{\circ}\) e cujas tangentes sejam conhecidas. Uma boa escolha é: - \( A = 45^{\circ} \) - \( B = 30^{\circ} \) ### Passo 2: Aplicar a Fórmula da Tangente da Soma A fórmula da tangente da soma de dois ângulos é: \[ \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \cdot \tan B} \] Substituindo \( A = 45^{\circ} \) e \( B = 30^{\circ} \): \[ \tan(75^{\circ}) = \frac{\tan(45^{\circ}) + \tan(30^{\circ})}{1 - \tan(45^{\circ}) \cdot \tan(30^{\circ})} \] ### Passo 3: Substituir os Valores das Tangentes Conhecidas Sabemos que: - \( \tan(45^{\circ}) = 1 \) - \( \tan(30^{\circ}) = \frac{\sqrt{3}}{3} \) Substituindo na equação: \[ \tan(75^{\circ}) = \frac{1 + \frac{\sqrt{3}}{3}}{1 - 1 \cdot \frac{\sqrt{3}}{3}} = \frac{\frac{3 + \sqrt{3}}{3}}{\frac{3 - \sqrt{3}}{3}} = \frac{3 + \sqrt{3}}{3 - \sqrt{3}} \] ### Passo 4: Racionalizar o Denominador Para simplificar a expressão, multiplicamos o numerador e o denominador pelo conjugado do denominador: \[ \frac{3 + \sqrt{3}}{3 - \sqrt{3}} \cdot \frac{3 + \sqrt{3}}{3 + \sqrt{3}} = \frac{(3 + \sqrt{3})^2}{(3)^2 - (\sqrt{3})^2} = \frac{9 + 6\sqrt{3} + 3}{9 - 3} = \frac{12 + 6\sqrt{3}}{6} \] Simplificando: \[ \frac{12 + 6\sqrt{3}}{6} = 2 + \sqrt{3} \] ### Resultado Final Portanto: \[ \tan(75^{\circ}) = 2 + \sqrt{3} \approx 3.732 \]

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Mind Expander

To calculate \( \tan(75^{\circ}) \) using the difference formula, we can express \( 75^{\circ} \) as \( 45^{\circ} + 30^{\circ} \). We can use the tangent addition formula: \[ \tan(A + B) = \frac{\tan A + \tan B}{1 - \tan A \tan B} \] Setting \( A = 45^{\circ} \) and \( B = 30^{\circ} \): 1. We know that \( \tan(45^{\circ}) = 1 \) and \( \tan(30^{\circ}) = \frac{1}{\sqrt{3}} \). 2. Plugging in the values, we have: \[ \tan(75^{\circ}) = \tan(45^{\circ} + 30^{\circ}) = \frac{\tan(45^{\circ}) + \tan(30^{\circ})}{1 - \tan(45^{\circ}) \tan(30^{\circ})} \] 3. Substituting the values: \[ \tan(75^{\circ}) = \frac{1 + \frac{1}{\sqrt{3}}}{1 - 1 \cdot \frac{1}{\sqrt{3}}} = \frac{1 + \frac{1}{\sqrt{3}}}{1 - \frac{1}{\sqrt{3}}} \] 4. To simplify this, we can multiply the numerator and denominator by \( \sqrt{3} \): \[ \tan(75^{\circ}) = \frac{\sqrt{3}(1 + \frac{1}{\sqrt{3}})}{\sqrt{3}(1 - \frac{1}{\sqrt{3}})} = \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \] 5. For further simplification, we can rationalize the denominator: \[ \tan(75^{\circ}) = \frac{(\sqrt{3} + 1)(\sqrt{3} + 1)}{(\sqrt{3} - 1)(\sqrt{3} + 1)} = \frac{(\sqrt{3} + 1)^2}{3 - 1} = \frac{3 + 2\sqrt{3} + 1}{2} = \frac{4 + 2\sqrt{3}}{2} = 2 + \sqrt{3} \] So, the final result is: \[ \tan(75^{\circ}) = 2 + \sqrt{3} \]

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