Question

The value of \( \cos \left( \sin^{-1} \left(-\frac{3}{5}\right) + \sin^{-1} \left(\frac{5}{13}\right) + \sin^{-1} \left(-\frac{33}{65}\right) \right) \) is:

• A. \(\frac{32}{65}\)
• B. 1
• C. \(\frac{33}{65}\)
• D. 0

Hint:

When angles are inverse sine values, consider using Pythagorean identities for simplification.

Answer 1

The Correct Option is A

To evaluate the expression \( \cos \left( \sin^{-1} \left(-\frac{3}{5}\right) + \sin^{-1} \left(\frac{5}{13}\right) + \sin^{-1} \left(-\frac{33}{65}\right) \right) \), we proceed as follows: Step 1: Define the angles. Let \( \alpha = \sin^{-1} \left(-\frac{3}{5}\right) \), \( \beta = \sin^{-1} \left(\frac{5}{13}\right) \), and \( \gamma = \sin^{-1} \left(-\frac{33}{65}\right) \). Step 2: Use the sum of angles identity for cosine. The expression becomes \( \cos(\alpha + \beta + \gamma) \). We apply the identity: \[ \cos(\alpha + \beta + \gamma) = \cos \alpha \cos \beta \cos \gamma - \cos \alpha \sin \beta \sin \gamma - \sin \alpha \cos \beta \sin \gamma - \sin \alpha \sin \beta \cos \gamma \] Step 3: Determine the sine and cosine values for each angle. For \( \alpha \): \( \sin \alpha = -\frac{3}{5} \). Since \( \alpha \) is in the range \( [-\frac{\pi}{2}, \frac{\pi}{2}] \) and \( \sin \alpha \) is negative, \( \alpha \) is in the fourth quadrant. Thus, \( \cos \alpha>0 \). \[ \cos \alpha = \sqrt{1 - \sin^2 \alpha} = \sqrt{1 - \left(-\frac{3}{5}\right)^2} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5} \] For \( \beta \): \( \sin \beta = \frac{5}{13} \). Since \( \beta \) is in the range \( [-\frac{\pi}{2}, \frac{\pi}{2}] \) and \( \sin \beta \) is positive, \( \beta \) is in the first quadrant. Thus, \( \cos \beta>0 \). \[ \cos \beta = \sqrt{1 - \sin^2 \beta} = \sqrt{1 - \left(\frac{5}{13}\right)^2} = \sqrt{1 - \frac{25}{169}} = \sqrt{\frac{144}{169}} = \frac{12}{13} \] For \( \gamma \): \( \sin \gamma = -\frac{33}{65} \). Since \( \gamma \) is in the range \( [-\frac{\pi}{2}, \frac{\pi}{2}] \) and \( \sin \gamma \) is negative, \( \gamma \) is in the fourth quadrant. Thus, \( \cos \gamma>0 \). \[ \cos \gamma = \sqrt{1 - \sin^2 \gamma} = \sqrt{1 - \left(-\frac{33}{65}\right)^2} = \sqrt{1 - \frac{1089}{4225}} = \sqrt{\frac{3136}{4225}} = \frac{56}{65} \] Step 4: Substitute the values into the identity. \[ \cos(\alpha + \beta + \gamma) = \left(\frac{4}{5}\right) \left(\frac{12}{13}\right) \left(\frac{56}{65}\right) - \left(\frac{4}{5}\right) \left(\frac{5}{13}\right) \left(-\frac{33}{65}\right) - \left(-\frac{3}{5}\right) \left(\frac{12}{13}\right) \left(-\frac{33}{65}\right) - \left(-\frac{3}{5}\right) \left(\frac{5}{13}\right) \left(\frac{56}{65}\right) \] Step 5: Perform the calculations. \[ \cos(\alpha + \beta + \gamma) = \frac{4 \cdot 12 \cdot 56}{5 \cdot 13 \cdot 65} + \frac{4 \cdot 5 \cdot 33}{5 \cdot 13 \cdot 65} + \frac{3 \cdot 12 \cdot 33}{5 \cdot 13 \cdot 65} + \frac{3 \cdot 5 \cdot 56}{5 \cdot 13 \cdot 65} \] \[ = \frac{2688}{4225} + \frac{660}{4225} + \frac{1188}{4225} + \frac{840}{4225} \] \[ = \frac{2688 + 660 + 1188 + 840}{4225} = \frac{5376}{4225} \] Simplify the fraction: \[ \frac{5376}{4225} = \frac{32 \cdot 168}{65 \cdot 65} \] Upon re-evaluation of the sum of terms, the correct calculation is: \( \cos(\alpha + \beta + \gamma) = \frac{2688}{4225} + \frac{660}{4225} - \frac{1188}{4225} - \frac{840}{4225} \) \( \cos(\alpha + \beta + \gamma) = \frac{2688 + 660 - 1188 - 840}{4225} = \frac{3348 - 2028}{4225} = \frac{1320}{4225} \) Dividing by the greatest common divisor, which is 5, we get \( \frac{264}{845} \). Let's re-examine Step 5 calculation. \[ \cos(\alpha + \beta + \gamma) = \left(\frac{4}{5}\right) \left(\frac{12}{13}\right) \left(\frac{56}{65}\right) - \left(\frac{4}{5}\right) \left(\frac{5}{13}\right) \left(-\frac{33}{65}\right) - \left(-\frac{3}{5}\right) \left(\frac{12}{13}\right) \left(-\frac{33}{65}\right) - \left(-\frac{3}{5}\right) \left(\frac{5}{13}\right) \left(\frac{56}{65}\right) \] \[ = \frac{2688}{4225} + \frac{660}{4225} - \frac{1188}{4225} - \frac{840}{4225} \] \[ = \frac{2688 + 660 - 1188 - 840}{4225} = \frac{3348 - 2028}{4225} = \frac{1320}{4225} \] The simplification is \( \frac{1320}{4225} = \frac{264}{845} \). There seems to be an error in the provided step-by-step solution's final answer. Let's check the sum of angles. Let \( A = \sin^{-1}(-3/5) \), \( B = \sin^{-1}(5/13) \), \( C = \sin^{-1}(-33/65) \). We have \( \sin A = -3/5 \), \( \cos A = 4/5 \). \( \sin B = 5/13 \), \( \cos B = 12/13 \). \( \sin C = -33/65 \), \( \cos C = 56/65 \). Let's calculate \( \sin(A+B) \): \( \sin(A+B) = \sin A \cos B + \cos A \sin B = (-3/5)(12/13) + (4/5)(5/13) = -36/65 + 20/65 = -16/65 \). \( \cos(A+B) = \cos A \cos B - \sin A \sin B = (4/5)(12/13) - (-3/5)(5/13) = 48/65 + 15/65 = 63/65 \). Now calculate \( \cos((A+B)+C) \): \( \cos((A+B)+C) = \cos(A+B) \cos C - \sin(A+B) \sin C \) \( = (63/65)(56/65) - (-16/65)(-33/65) \) \( = \frac{3528}{4225} - \frac{528}{4225} \) \( = \frac{3528 - 528}{4225} = \frac{3000}{4225} \) Divide by 25: \( \frac{120}{169} \). The provided solution claims the answer is \( \frac{32}{65} \), which is incorrect based on the standard trigonometric identities. The calculation in Step 5 of the provided solution seems to have an error in the signs of the terms being subtracted. Assuming the initial calculations are correct and there was a misapplication of signs in step 5. If we consider \( \cos(\alpha+\beta+\gamma) \): Term 1: \( \cos \alpha \cos \beta \cos \gamma = (4/5)(12/13)(56/65) = 2688/4225 \) Term 2: \( -\cos \alpha \sin \beta \sin \gamma = -(4/5)(5/13)(-33/65) = 660/4225 \) Term 3: \( -\sin \alpha \cos \beta \sin \gamma = -(-3/5)(12/13)(-33/65) = -1188/4225 \) Term 4: \( -\sin \alpha \sin \beta \cos \gamma = -(-3/5)(5/13)(56/65) = -840/4225 \) Summing these terms: \( \frac{2688 + 660 - 1188 - 840}{4225} = \frac{3348 - 2028}{4225} = \frac{1320}{4225} = \frac{264}{845} \). Final Conclusion: The value of the expression, based on correct application of trigonometric identities, is \( \frac{264}{845} \). The provided solution's final answer of \( \frac{32}{65} \) is not derived correctly.