Question

If \( a \) is in the 3rd quadrant, \( \beta \) is in the 2nd quadrant such that \( \tan \alpha = \frac{1}{7}, \sin \beta = \frac{1}{\sqrt{10}} \), then \[ \sin(2\alpha + \beta) = \]

• A. \( \frac{3 \times \sqrt{10}}{25} \)
• B. \( \frac{3}{\sqrt{10}} \)
• C. \( \frac{3}{25\sqrt{10}} \)
• D. \( \frac{\sqrt{10}}{3 \times 25} \)

Hint:

For trigonometric expressions involving multiple angles, use the angle addition and double angle formulas to express the function in terms of simpler trigonometric functions. Then simplify using known identities.

Answer 1

The Correct Option is C

Step 1: Find \( \sin(\alpha) \) and \( \cos(\alpha) \). Given \( \tan(\alpha) = \frac{1}{7} \), we calculate \( \sin(\alpha) \) and \( \cos(\alpha) \) using the identity \( \tan^2(\alpha) + 1 = \sec^2(\alpha) \).

Step 2: Find \( \sin(\beta) \) and \( \cos(\beta) \). Given \( \sin(\beta) = \frac{1}{\sqrt{10}} \), we calculate \( \cos(\beta) \) using the identity \( \sin^2(\beta) + \cos^2(\beta) = 1 \).

Step 3: Apply the angle addition formula for \( \sin(2\alpha + \beta) \). We use the identity \( \sin(2\alpha + \beta) = \sin(2\alpha) \cos(\beta) + \cos(2\alpha) \sin(\beta) \), and the double angle formulas for sine and cosine to calculate the value of \( \sin(2\alpha + \beta) \).

Step 4: Final result. The final result is: \[ \sin(2\alpha + \beta) = \frac{3 \times \sqrt{10}}{25}. \]

Answer 2

Given: - \( \alpha \) lies in the 3rd quadrant with \( \tan \alpha = \frac{1}{7} \).
- \( \beta \) lies in the 2nd quadrant with \( \sin \beta = \frac{1}{\sqrt{10}} \).

Find: \[ \sin(2\alpha + \beta) \]

Step 1: Find \(\sin \alpha\) and \(\cos \alpha\)
Since \( \tan \alpha = \frac{1}{7} \), let the opposite side = 1 and adjacent side = 7. The hypotenuse is \[ \sqrt{1^2 + 7^2} = \sqrt{50} = 5\sqrt{2}. \] Because \( \alpha \) is in the 3rd quadrant, both sine and cosine are negative: \[ \sin \alpha = -\frac{1}{5\sqrt{2}}, \quad \cos \alpha = -\frac{7}{5\sqrt{2}}. \]

Step 2: Find \(\cos \beta\)
Given \( \sin \beta = \frac{1}{\sqrt{10}} \), and \( \beta \) is in the 2nd quadrant where cosine is negative: \[ \cos \beta = -\sqrt{1 - \sin^2 \beta} = -\sqrt{1 - \frac{1}{10}} = -\sqrt{\frac{9}{10}} = -\frac{3}{\sqrt{10}}. \]

Step 3: Calculate \(\sin 2\alpha\) and \(\cos 2\alpha\)
Using double angle formulas: \[ \sin 2\alpha = 2 \sin \alpha \cos \alpha = 2 \times \left(-\frac{1}{5\sqrt{2}}\right) \times \left(-\frac{7}{5\sqrt{2}}\right) = 2 \times \frac{7}{50} = \frac{14}{50} = \frac{7}{25}. \] \[ \cos 2\alpha = \cos^2 \alpha - \sin^2 \alpha = \left(-\frac{7}{5\sqrt{2}}\right)^2 - \left(-\frac{1}{5\sqrt{2}}\right)^2 = \frac{49}{50} - \frac{1}{50} = \frac{48}{50} = \frac{24}{25}. \]

Step 4: Use angle addition formula for sine
\[ \sin(2\alpha + \beta) = \sin 2\alpha \cos \beta + \cos 2\alpha \sin \beta. \] Substitute values: \[ = \frac{7}{25} \times \left(-\frac{3}{\sqrt{10}}\right) + \frac{24}{25} \times \frac{1}{\sqrt{10}} = -\frac{21}{25 \sqrt{10}} + \frac{24}{25 \sqrt{10}} = \frac{3}{25 \sqrt{10}}. \]

Final answer: \[ \boxed{\frac{3}{25 \sqrt{10}}}. \]