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Simplify the Sollowing D \( \sin \left(\theta-45^{\circ}\right)+\sin \left(\theta+45^{\circ}\right. \)

Ask by Salazar Ramos. in South Africa
Feb 03,2025

Real Tutor Solution

Tutor-Verified Answer

Answer

The simplified expression is \( \sqrt{2} \sin(\theta) \).

Solution

Simplify the expression by following steps: - step0: Solution: \(\sin\left(\theta -45\right)+\sin\left(\theta +45\right)\) - step1: Transform the expression: \(2\sin\left(\theta \right)\cos\left(-45\right)\) - step2: Calculate: \(2\sin\left(\theta \right)\cos\left(45\right)\) - step3: Multiply the terms: \(2\cos\left(45\right)\sin\left(\theta \right)\) To simplify the expression \( \sin(\theta - 45^{\circ}) + \sin(\theta + 45^{\circ}) \), we can use the sine addition and subtraction formulas. 1. **Apply the sine addition and subtraction formulas:** - The formula for \( \sin(a \pm b) \) is: \[ \sin(a \pm b) = \sin(a)\cos(b) \pm \cos(a)\sin(b) \] - For \( a = \theta \) and \( b = 45^{\circ} \): - \( \sin(\theta - 45^{\circ}) = \sin(\theta)\cos(45^{\circ}) - \cos(\theta)\sin(45^{\circ}) \) - \( \sin(\theta + 45^{\circ}) = \sin(\theta)\cos(45^{\circ}) + \cos(\theta)\sin(45^{\circ}) \) 2. **Combine the two expressions:** \[ \sin(\theta - 45^{\circ}) + \sin(\theta + 45^{\circ}) = \left(\sin(\theta)\cos(45^{\circ}) - \cos(\theta)\sin(45^{\circ})\right) + \left(\sin(\theta)\cos(45^{\circ}) + \cos(\theta)\sin(45^{\circ})\right) \] 3. **Simplify:** - The terms involving \( \cos(\theta)\sin(45^{\circ}) \) cancel out: \[ = 2\sin(\theta)\cos(45^{\circ}) \] 4. **Substituting the value of \( \cos(45^{\circ}) \):** - Since \( \cos(45^{\circ}) = \frac{\sqrt{2}}{2} \): \[ = 2 \cdot \frac{\sqrt{2}}{2} \sin(\theta) = \sqrt{2} \sin(\theta) \] Thus, the simplified expression is: \[ \sqrt{2} \sin(\theta) \]

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Mind Expander

To simplify the expression \( D \sin \left(\theta - 45^{\circ}\right) + \sin \left(\theta + 45^{\circ}\right) \), we can use the sine addition and subtraction formulas. First, recall the formulas: - \( \sin(A - B) = \sin A \cos B - \cos A \sin B \) - \( \sin(A + B) = \sin A \cos B + \cos A \sin B \) Applying these to our problem: 1. For \( \sin(\theta - 45^{\circ}) \): \[ \sin(\theta - 45^{\circ}) = \sin \theta \cos 45^{\circ} - \cos \theta \sin 45^{\circ} = \sin \theta \cdot \frac{\sqrt{2}}{2} - \cos \theta \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{2} (\sin \theta - \cos \theta) \] 2. For \( \sin(\theta + 45^{\circ}) \): \[ \sin(\theta + 45^{\circ}) = \sin \theta \cos 45^{\circ} + \cos \theta \sin 45^{\circ} = \sin \theta \cdot \frac{\sqrt{2}}{2} + \cos \theta \cdot \frac{\sqrt{2}}{2} = \frac{\sqrt{2}}{2} (\sin \theta + \cos \theta) \] Now substituting these back into the original expression: \[ D \sin(\theta - 45^{\circ}) + \sin(\theta + 45^{\circ}) = D \cdot \frac{\sqrt{2}}{2} (\sin \theta - \cos \theta) + \frac{\sqrt{2}}{2} (\sin \theta + \cos \theta) \] Factoring out \( \frac{\sqrt{2}}{2} \): \[ = \frac{\sqrt{2}}{2} \left( D (\sin \theta - \cos \theta) + (\sin \theta + \cos \theta) \right) \] \[ = \frac{\sqrt{2}}{2} \left( (D + 1) \sin \theta + (-D + 1) \cos \theta \right) \] So the simplified expression is: \[ \frac{\sqrt{2}}{2} \left( (D + 1) \sin \theta + (1 - D) \cos \theta \right) \]

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