If \( \cos \alpha=0.689 \) and \( \sin \beta=0.026 \) with botb angles' terminal rays in Quadrant-1, find the values of (a) \( \cos (\beta-\alpha)=\square \) (b) \( \sin (\alpha-\beta)=\square \) Your answers should be accurate to 4 decimal places. Submit Question

To find the values of \( \cos (\beta - \alpha) \) and \( \sin (\alpha - \beta) \), we can use the cosine and sine subtraction formulas. First, recall the formulas: 1. \( \cos (\beta - \alpha) = \cos \beta \cos \alpha + \sin \beta \sin \alpha \) 2. \( \sin (\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta \) Given: - \( \cos \alpha = 0.689 \) - \( \sin \beta = 0.026 \) We need to find \( \sin \alpha \) and \( \cos \beta \). Using the Pythagorean identity: \[ \sin^2 \alpha + \cos^2 \alpha = 1 \] We find \( \sin \alpha \): \[ \sin \alpha = \sqrt{1 - \cos^2 \alpha} = \sqrt{1 - (0.689)^2} = \sqrt{1 - 0.474721} = \sqrt{0.525279} \approx 0.7255 \] Using the same identity for \( \sin \beta \) to find \( \cos \beta \): \[ \cos^2 \beta + \sin^2 \beta = 1 \] \[ \cos \beta = \sqrt{1 - \sin^2 \beta} = \sqrt{1 - (0.026)^2} = \sqrt{1 - 0.000676} = \sqrt{0.999324} \approx 0.9997 \] Now we can find \( \cos (\beta - \alpha) \): \[ \cos (\beta - \alpha) = \cos \beta \cos \alpha + \sin \beta \sin \alpha \] \[ = (0.9997)(0.689) + (0.026)(0.7255) \approx 0.6887 + 0.0188 = 0.7075 \] Next, we calculate \( \sin (\alpha - \beta) \): \[ \sin (\alpha - \beta) = \sin \alpha \cos \beta - \cos \alpha \sin \beta \] \[ = (0.7255)(0.9997) - (0.689)(0.026) \approx 0.7253 - 0.0179 = 0.7074 \] Thus, the final answers are: (a) \( \cos (\beta - \alpha) \approx 0.7075 \) (b) \( \sin (\alpha - \beta) \approx 0.7074 \) To four decimal places: (a) \( \cos (\beta - \alpha) = 0.7075 \) (b) \( \sin (\alpha - \beta) = 0.7074 \)