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Sagot :
Certainly! Let's go through the problem step-by-step.
Given:
- [tex]\(\sin \alpha = \frac{4}{5}\)[/tex] with [tex]\(\alpha\)[/tex] in Quadrant II
- [tex]\(\cos \beta = \frac{2}{5}\)[/tex] with [tex]\(\beta\)[/tex] in Quadrant I
We need to find [tex]\(\cos (\alpha - \beta)\)[/tex].
Step 1: Find [tex]\(\cos \alpha\)[/tex]
Since [tex]\(\alpha\)[/tex] is in Quadrant II, [tex]\(\cos \alpha\)[/tex] is negative. We can use the Pythagorean identity to find [tex]\(\cos \alpha\)[/tex]:
[tex]\[ \sin^2 \alpha + \cos^2 \alpha = 1 \][/tex]
Plugging in the given value for [tex]\(\sin \alpha\)[/tex]:
[tex]\[ \left(\frac{4}{5}\right)^2 + \cos^2 \alpha = 1 \][/tex]
[tex]\[ \frac{16}{25} + \cos^2 \alpha = 1 \][/tex]
[tex]\[ \cos^2 \alpha = 1 - \frac{16}{25} \][/tex]
[tex]\[ \cos^2 \alpha = \frac{25}{25} - \frac{16}{25} \][/tex]
[tex]\[ \cos^2 \alpha = \frac{9}{25} \][/tex]
Since [tex]\(\alpha\)[/tex] is in Quadrant II, [tex]\(\cos \alpha\)[/tex] is negative:
[tex]\[ \cos \alpha = -\sqrt{\frac{9}{25}} = -\frac{3}{5} \][/tex]
Step 2: Find [tex]\(\sin \beta\)[/tex]
Since [tex]\(\beta\)[/tex] is in Quadrant I, [tex]\(\sin \beta\)[/tex] is positive. Using the Pythagorean identity:
[tex]\[ \sin^2 \beta + \cos^2 \beta = 1 \][/tex]
Plugging in the given value for [tex]\(\cos \beta\)[/tex]:
[tex]\[ \sin^2 \beta + \left(\frac{2}{5}\right)^2 = 1 \][/tex]
[tex]\[ \sin^2 \beta + \frac{4}{25} = 1 \][/tex]
[tex]\[ \sin^2 \beta = 1 - \frac{4}{25} \][/tex]
[tex]\[ \sin^2 \beta = \frac{25}{25} - \frac{4}{25} \][/tex]
[tex]\[ \sin^2 \beta = \frac{21}{25} \][/tex]
Since [tex]\(\beta\)[/tex] is in Quadrant I, [tex]\(\sin \beta\)[/tex] is positive:
[tex]\[ \sin \beta = \sqrt{\frac{21}{25}} = \frac{\sqrt{21}}{5} \][/tex]
Step 3: Calculate [tex]\(\cos (\alpha - \beta)\)[/tex]
Using the cosine subtraction formula:
[tex]\[ \cos (\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta \][/tex]
Substitute the values calculated:
[tex]\[ \cos (\alpha - \beta) = \left(-\frac{3}{5}\right) \left(\frac{2}{5}\right) + \left(\frac{4}{5}\right) \left(\frac{\sqrt{21}}{5}\right) \][/tex]
[tex]\[ \cos (\alpha - \beta) = -\frac{6}{25} + \frac{4\sqrt{21}}{25} \][/tex]
[tex]\[ \cos (\alpha - \beta) = -\frac{6}{25} + \frac{4\sqrt{21}}{25} = \frac{-6 + 4\sqrt{21}}{25} \][/tex]
The values we had earlier compute nicely to [tex]\(0.4932121111929344\)[/tex], meaning:
[tex]\[ \cos (\alpha - \beta) \approx 0.4932 \text{ (approx)} \][/tex]
Given:
- [tex]\(\sin \alpha = \frac{4}{5}\)[/tex] with [tex]\(\alpha\)[/tex] in Quadrant II
- [tex]\(\cos \beta = \frac{2}{5}\)[/tex] with [tex]\(\beta\)[/tex] in Quadrant I
We need to find [tex]\(\cos (\alpha - \beta)\)[/tex].
Step 1: Find [tex]\(\cos \alpha\)[/tex]
Since [tex]\(\alpha\)[/tex] is in Quadrant II, [tex]\(\cos \alpha\)[/tex] is negative. We can use the Pythagorean identity to find [tex]\(\cos \alpha\)[/tex]:
[tex]\[ \sin^2 \alpha + \cos^2 \alpha = 1 \][/tex]
Plugging in the given value for [tex]\(\sin \alpha\)[/tex]:
[tex]\[ \left(\frac{4}{5}\right)^2 + \cos^2 \alpha = 1 \][/tex]
[tex]\[ \frac{16}{25} + \cos^2 \alpha = 1 \][/tex]
[tex]\[ \cos^2 \alpha = 1 - \frac{16}{25} \][/tex]
[tex]\[ \cos^2 \alpha = \frac{25}{25} - \frac{16}{25} \][/tex]
[tex]\[ \cos^2 \alpha = \frac{9}{25} \][/tex]
Since [tex]\(\alpha\)[/tex] is in Quadrant II, [tex]\(\cos \alpha\)[/tex] is negative:
[tex]\[ \cos \alpha = -\sqrt{\frac{9}{25}} = -\frac{3}{5} \][/tex]
Step 2: Find [tex]\(\sin \beta\)[/tex]
Since [tex]\(\beta\)[/tex] is in Quadrant I, [tex]\(\sin \beta\)[/tex] is positive. Using the Pythagorean identity:
[tex]\[ \sin^2 \beta + \cos^2 \beta = 1 \][/tex]
Plugging in the given value for [tex]\(\cos \beta\)[/tex]:
[tex]\[ \sin^2 \beta + \left(\frac{2}{5}\right)^2 = 1 \][/tex]
[tex]\[ \sin^2 \beta + \frac{4}{25} = 1 \][/tex]
[tex]\[ \sin^2 \beta = 1 - \frac{4}{25} \][/tex]
[tex]\[ \sin^2 \beta = \frac{25}{25} - \frac{4}{25} \][/tex]
[tex]\[ \sin^2 \beta = \frac{21}{25} \][/tex]
Since [tex]\(\beta\)[/tex] is in Quadrant I, [tex]\(\sin \beta\)[/tex] is positive:
[tex]\[ \sin \beta = \sqrt{\frac{21}{25}} = \frac{\sqrt{21}}{5} \][/tex]
Step 3: Calculate [tex]\(\cos (\alpha - \beta)\)[/tex]
Using the cosine subtraction formula:
[tex]\[ \cos (\alpha - \beta) = \cos \alpha \cos \beta + \sin \alpha \sin \beta \][/tex]
Substitute the values calculated:
[tex]\[ \cos (\alpha - \beta) = \left(-\frac{3}{5}\right) \left(\frac{2}{5}\right) + \left(\frac{4}{5}\right) \left(\frac{\sqrt{21}}{5}\right) \][/tex]
[tex]\[ \cos (\alpha - \beta) = -\frac{6}{25} + \frac{4\sqrt{21}}{25} \][/tex]
[tex]\[ \cos (\alpha - \beta) = -\frac{6}{25} + \frac{4\sqrt{21}}{25} = \frac{-6 + 4\sqrt{21}}{25} \][/tex]
The values we had earlier compute nicely to [tex]\(0.4932121111929344\)[/tex], meaning:
[tex]\[ \cos (\alpha - \beta) \approx 0.4932 \text{ (approx)} \][/tex]
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