Find the best solutions to your questions at Westonci.ca, the premier Q&A platform with a community of knowledgeable experts. Discover detailed solutions to your questions from a wide network of experts on our comprehensive Q&A platform. Get detailed and accurate answers to your questions from a dedicated community of experts on our Q&A platform.
Sagot :
Alright, let's tackle this problem step-by-step.
Given:
[tex]\[ \cos \left(\theta + 30^\circ\right) = \frac{1}{2} \sin \theta \][/tex]
We will use the angle addition formula for the cosine function:
[tex]\[ \cos \left(\theta + 30^\circ\right) = \cos \theta \cos 30^\circ - \sin \theta \sin 30^\circ \][/tex]
We know from trigonometric identities that:
[tex]\[ \cos 30^\circ = \frac{\sqrt{3}}{2} \quad \text{and} \quad \sin 30^\circ = \frac{1}{2} \][/tex]
Substituting these values into the angle addition formula, we get:
[tex]\[ \cos \left(\theta + 30^\circ\right) = \cos \theta \cdot \frac{\sqrt{3}}{2} - \sin \theta \cdot \frac{1}{2} \][/tex]
Let's substitute this back into the given equation:
[tex]\[ \cos \theta \cdot \frac{\sqrt{3}}{2} - \sin \theta \cdot \frac{1}{2} = \frac{1}{2} \sin \theta \][/tex]
Now, we need to simplify this equation:
[tex]\[ \cos \theta \cdot \frac{\sqrt{3}}{2} - \sin \theta \cdot \frac{1}{2} - \frac{1}{2} \sin \theta = 0 \][/tex]
Combine the [tex]\(\sin \theta\)[/tex] terms:
[tex]\[ \cos \theta \cdot \frac{\sqrt{3}}{2} - \sin \theta \left(\frac{1}{2} + \frac{1}{2}\right) = 0 \][/tex]
This simplifies to:
[tex]\[ \cos \theta \cdot \frac{\sqrt{3}}{2} - \sin \theta \cdot 1 = 0 \][/tex]
or:
[tex]\[ \cos \theta \cdot \frac{\sqrt{3}}{2} = \sin \theta \][/tex]
Rearranging the terms gives:
[tex]\[ \cos \theta \cdot \frac{\sqrt{3}}{2} = \sin \theta \][/tex]
To isolate [tex]\(\tan \theta\)[/tex], divide both sides by [tex]\(\cos \theta\)[/tex]:
[tex]\[ \frac{\sqrt{3}}{2} = \frac{\sin \theta}{\cos \theta} \][/tex]
Recall that [tex]\(\frac{\sin \theta}{\cos \theta} = \tan \theta\)[/tex]:
So, we have:
[tex]\[ \tan \theta = \frac{\sqrt{3}}{2} \][/tex]
Hence, we have shown that:
[tex]\[ \tan \theta = \frac{\sqrt{3}}{2} \][/tex]
This completes the solution.
Given:
[tex]\[ \cos \left(\theta + 30^\circ\right) = \frac{1}{2} \sin \theta \][/tex]
We will use the angle addition formula for the cosine function:
[tex]\[ \cos \left(\theta + 30^\circ\right) = \cos \theta \cos 30^\circ - \sin \theta \sin 30^\circ \][/tex]
We know from trigonometric identities that:
[tex]\[ \cos 30^\circ = \frac{\sqrt{3}}{2} \quad \text{and} \quad \sin 30^\circ = \frac{1}{2} \][/tex]
Substituting these values into the angle addition formula, we get:
[tex]\[ \cos \left(\theta + 30^\circ\right) = \cos \theta \cdot \frac{\sqrt{3}}{2} - \sin \theta \cdot \frac{1}{2} \][/tex]
Let's substitute this back into the given equation:
[tex]\[ \cos \theta \cdot \frac{\sqrt{3}}{2} - \sin \theta \cdot \frac{1}{2} = \frac{1}{2} \sin \theta \][/tex]
Now, we need to simplify this equation:
[tex]\[ \cos \theta \cdot \frac{\sqrt{3}}{2} - \sin \theta \cdot \frac{1}{2} - \frac{1}{2} \sin \theta = 0 \][/tex]
Combine the [tex]\(\sin \theta\)[/tex] terms:
[tex]\[ \cos \theta \cdot \frac{\sqrt{3}}{2} - \sin \theta \left(\frac{1}{2} + \frac{1}{2}\right) = 0 \][/tex]
This simplifies to:
[tex]\[ \cos \theta \cdot \frac{\sqrt{3}}{2} - \sin \theta \cdot 1 = 0 \][/tex]
or:
[tex]\[ \cos \theta \cdot \frac{\sqrt{3}}{2} = \sin \theta \][/tex]
Rearranging the terms gives:
[tex]\[ \cos \theta \cdot \frac{\sqrt{3}}{2} = \sin \theta \][/tex]
To isolate [tex]\(\tan \theta\)[/tex], divide both sides by [tex]\(\cos \theta\)[/tex]:
[tex]\[ \frac{\sqrt{3}}{2} = \frac{\sin \theta}{\cos \theta} \][/tex]
Recall that [tex]\(\frac{\sin \theta}{\cos \theta} = \tan \theta\)[/tex]:
So, we have:
[tex]\[ \tan \theta = \frac{\sqrt{3}}{2} \][/tex]
Hence, we have shown that:
[tex]\[ \tan \theta = \frac{\sqrt{3}}{2} \][/tex]
This completes the solution.
Thanks for stopping by. We are committed to providing the best answers for all your questions. See you again soon. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Thank you for visiting Westonci.ca, your go-to source for reliable answers. Come back soon for more expert insights.