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
\[
\sin(2\alpha + \beta) =
\]
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
\[
\sin(2\alpha + \beta) =
\]
Show 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.
Updated On: Mar 18, 2026
- \( \frac{3 \times \sqrt{10}}{25} \)
- \( \frac{3}{\sqrt{10}} \)
- \( \frac{3}{25\sqrt{10}} \)
- \( \frac{\sqrt{10}}{3 \times 25} \)
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The Correct Option is C
Approach Solution - 1
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: \[ \sin(2\alpha + \beta) = \frac{3 \times \sqrt{10}}{25}. \]
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Approach Solution -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: \[ \boxed{\frac{3}{25 \sqrt{10}}}. \]
- \( \beta \) 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: \[ \boxed{\frac{3}{25 \sqrt{10}}}. \]
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