Westonci.ca connects you with experts who provide insightful answers to your questions. Join us today and start learning! Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately. Get quick and reliable solutions to your questions from a community of experienced experts on our platform.
Sagot :
To solve for \(\cos (\alpha + \beta)\), \(\sin (\alpha + \beta)\), and \(\tan (\alpha + \beta)\) given \(\sin (\alpha)=\frac{4}{5}\) where \(\alpha\) lies in quadrant I, and \(\sin (\beta)=\frac{24}{25}\) where \(\beta\) lies in quadrant II, let's proceed step-by-step.
### Step 1: Finding \(\cos(\alpha)\)
Since \(\alpha\) is in quadrant I, \(\cos(\alpha)\) will be positive.
We have:
[tex]\[ \sin(\alpha) = \frac{4}{5} \][/tex]
Using the Pythagorean identity \(\sin^2(\alpha) + \cos^2(\alpha) = 1\):
[tex]\[ \cos^2(\alpha) = 1 - \sin^2(\alpha) \][/tex]
[tex]\[ \cos^2(\alpha) = 1 - \left( \frac{4}{5} \right)^2 \][/tex]
[tex]\[ \cos^2(\alpha) = 1 - \frac{16}{25} \][/tex]
[tex]\[ \cos^2(\alpha) = \frac{9}{25} \][/tex]
[tex]\[ \cos(\alpha) = \sqrt{\frac{9}{25}} \][/tex]
[tex]\[ \cos(\alpha) = \frac{3}{5} \][/tex]
### Step 2: Finding \(\cos(\beta)\)
Since \(\beta\) is in quadrant II, \(\cos(\beta)\) will be negative.
We have:
[tex]\[ \sin(\beta) = \frac{24}{25} \][/tex]
Using the Pythagorean identity \(\sin^2(\beta) + \cos^2(\beta) = 1\):
[tex]\[ \cos^2(\beta) = 1 - \sin^2(\beta) \][/tex]
[tex]\[ \cos^2(\beta) = 1 - \left( \frac{24}{25} \right)^2 \][/tex]
[tex]\[ \cos^2(\beta) = 1 - \frac{576}{625} \][/tex]
[tex]\[ \cos^2(\beta) = \frac{49}{625} \][/tex]
[tex]\[ \cos(\beta) = -\sqrt{\frac{49}{625}} \][/tex]
[tex]\[ \cos(\beta) = -\frac{7}{25} \][/tex]
### Step 3: Using Angle Addition Formulas
#### Finding \(\cos (\alpha + \beta)\)
[tex]\[ \cos (\alpha + \beta) = \cos(\alpha) \cos(\beta) - \sin(\alpha) \sin(\beta) \][/tex]
Substitute the values for \(\cos(\alpha)\), \(\cos(\beta)\), \(\sin(\alpha)\), and \(\sin(\beta)\):
[tex]\[ \cos (\alpha + \beta) = \left( \frac{3}{5} \right) \left( -\frac{7}{25} \right) - \left( \frac{4}{5} \right) \left( \frac{24}{25} \right) \][/tex]
[tex]\[ \cos (\alpha + \beta) = \left( \frac{3 \times -7}{125} \right) - \left( \frac{4 \times 24}{125} \right) \][/tex]
[tex]\[ \cos (\alpha + \beta) = \frac{-21}{125} - \frac{96}{125} \][/tex]
[tex]\[ \cos (\alpha + \beta) = \frac{-117}{125} \][/tex]
[tex]\[ \cos (\alpha + \beta) = -0.936 \][/tex]
So,
[tex]\[ \cos (\alpha + \beta) = - \boxed{0.936} \][/tex]
#### Finding \(\sin (\alpha + \beta)\)
[tex]\[ \sin (\alpha + \beta) = \sin(\alpha) \cos(\beta) + \cos(\alpha) \sin(\beta) \][/tex]
Substitute the values for \(\cos(\alpha)\), \(\cos(\beta)\), \(\sin(\alpha)\), and \(\sin(\beta)\):
[tex]\[ \sin (\alpha + \beta) = \left( \frac{4}{5} \right) \left( -\frac{7}{25} \right) + \left( \frac{3}{5} \right) \left( \frac{24}{25} \right) \][/tex]
[tex]\[ \sin (\alpha + \beta) = \left( \frac{4 \times -7}{125} \right) + \left( \frac{3 \times 24}{125} \right) \][/tex]
[tex]\[ \sin (\alpha + \beta) = \frac{-28}{125} + \frac{72}{125} \][/tex]
[tex]\[ \sin (\alpha + \beta) = \frac{44}{125} \][/tex]
[tex]\[ \sin (\alpha + \beta) = 0.352 \][/tex]
So,
[tex]\[ \sin (\alpha + \beta) = \boxed{0.352} \][/tex]
#### Finding \(\tan (\alpha + \beta)\)
[tex]\[ \tan (\alpha + \beta) = \frac{\sin (\alpha + \beta)}{\cos (\alpha + \beta)} \][/tex]
Substitute the values for \(\sin(\alpha + \beta)\) and \(\cos(\alpha + \beta)\):
[tex]\[ \tan (\alpha + \beta) = \frac{0.352}{-0.936} \][/tex]
[tex]\[ \tan (\alpha + \beta) = -0.376 \][/tex]
So,
[tex]\[ \tan (\alpha + \beta) = - \boxed{0.376} \][/tex]
These values match the results obtained from calculations:
[tex]\[ \cos (\alpha + \beta) = -0.936 \][/tex]
[tex]\[ \sin (\alpha + \beta) = 0.352 \][/tex]
[tex]\[ \tan (\alpha + \beta) = -0.376 \][/tex]
### Step 1: Finding \(\cos(\alpha)\)
Since \(\alpha\) is in quadrant I, \(\cos(\alpha)\) will be positive.
We have:
[tex]\[ \sin(\alpha) = \frac{4}{5} \][/tex]
Using the Pythagorean identity \(\sin^2(\alpha) + \cos^2(\alpha) = 1\):
[tex]\[ \cos^2(\alpha) = 1 - \sin^2(\alpha) \][/tex]
[tex]\[ \cos^2(\alpha) = 1 - \left( \frac{4}{5} \right)^2 \][/tex]
[tex]\[ \cos^2(\alpha) = 1 - \frac{16}{25} \][/tex]
[tex]\[ \cos^2(\alpha) = \frac{9}{25} \][/tex]
[tex]\[ \cos(\alpha) = \sqrt{\frac{9}{25}} \][/tex]
[tex]\[ \cos(\alpha) = \frac{3}{5} \][/tex]
### Step 2: Finding \(\cos(\beta)\)
Since \(\beta\) is in quadrant II, \(\cos(\beta)\) will be negative.
We have:
[tex]\[ \sin(\beta) = \frac{24}{25} \][/tex]
Using the Pythagorean identity \(\sin^2(\beta) + \cos^2(\beta) = 1\):
[tex]\[ \cos^2(\beta) = 1 - \sin^2(\beta) \][/tex]
[tex]\[ \cos^2(\beta) = 1 - \left( \frac{24}{25} \right)^2 \][/tex]
[tex]\[ \cos^2(\beta) = 1 - \frac{576}{625} \][/tex]
[tex]\[ \cos^2(\beta) = \frac{49}{625} \][/tex]
[tex]\[ \cos(\beta) = -\sqrt{\frac{49}{625}} \][/tex]
[tex]\[ \cos(\beta) = -\frac{7}{25} \][/tex]
### Step 3: Using Angle Addition Formulas
#### Finding \(\cos (\alpha + \beta)\)
[tex]\[ \cos (\alpha + \beta) = \cos(\alpha) \cos(\beta) - \sin(\alpha) \sin(\beta) \][/tex]
Substitute the values for \(\cos(\alpha)\), \(\cos(\beta)\), \(\sin(\alpha)\), and \(\sin(\beta)\):
[tex]\[ \cos (\alpha + \beta) = \left( \frac{3}{5} \right) \left( -\frac{7}{25} \right) - \left( \frac{4}{5} \right) \left( \frac{24}{25} \right) \][/tex]
[tex]\[ \cos (\alpha + \beta) = \left( \frac{3 \times -7}{125} \right) - \left( \frac{4 \times 24}{125} \right) \][/tex]
[tex]\[ \cos (\alpha + \beta) = \frac{-21}{125} - \frac{96}{125} \][/tex]
[tex]\[ \cos (\alpha + \beta) = \frac{-117}{125} \][/tex]
[tex]\[ \cos (\alpha + \beta) = -0.936 \][/tex]
So,
[tex]\[ \cos (\alpha + \beta) = - \boxed{0.936} \][/tex]
#### Finding \(\sin (\alpha + \beta)\)
[tex]\[ \sin (\alpha + \beta) = \sin(\alpha) \cos(\beta) + \cos(\alpha) \sin(\beta) \][/tex]
Substitute the values for \(\cos(\alpha)\), \(\cos(\beta)\), \(\sin(\alpha)\), and \(\sin(\beta)\):
[tex]\[ \sin (\alpha + \beta) = \left( \frac{4}{5} \right) \left( -\frac{7}{25} \right) + \left( \frac{3}{5} \right) \left( \frac{24}{25} \right) \][/tex]
[tex]\[ \sin (\alpha + \beta) = \left( \frac{4 \times -7}{125} \right) + \left( \frac{3 \times 24}{125} \right) \][/tex]
[tex]\[ \sin (\alpha + \beta) = \frac{-28}{125} + \frac{72}{125} \][/tex]
[tex]\[ \sin (\alpha + \beta) = \frac{44}{125} \][/tex]
[tex]\[ \sin (\alpha + \beta) = 0.352 \][/tex]
So,
[tex]\[ \sin (\alpha + \beta) = \boxed{0.352} \][/tex]
#### Finding \(\tan (\alpha + \beta)\)
[tex]\[ \tan (\alpha + \beta) = \frac{\sin (\alpha + \beta)}{\cos (\alpha + \beta)} \][/tex]
Substitute the values for \(\sin(\alpha + \beta)\) and \(\cos(\alpha + \beta)\):
[tex]\[ \tan (\alpha + \beta) = \frac{0.352}{-0.936} \][/tex]
[tex]\[ \tan (\alpha + \beta) = -0.376 \][/tex]
So,
[tex]\[ \tan (\alpha + \beta) = - \boxed{0.376} \][/tex]
These values match the results obtained from calculations:
[tex]\[ \cos (\alpha + \beta) = -0.936 \][/tex]
[tex]\[ \sin (\alpha + \beta) = 0.352 \][/tex]
[tex]\[ \tan (\alpha + \beta) = -0.376 \][/tex]
Visit us again for up-to-date and reliable answers. We're always ready to assist you with your informational needs. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Thank you for visiting Westonci.ca. Stay informed by coming back for more detailed answers.