Welcome to Westonci.ca, where curiosity meets expertise. Ask any question and receive fast, accurate answers from our knowledgeable community. Join our Q&A platform and connect with professionals ready to provide precise answers to your questions in various areas. Experience the convenience of finding accurate answers to your questions from knowledgeable experts on our platform.
Sagot :
Certainly! Let's solve the given problem step-by-step.
Given:
[tex]\[ \cos(\theta) = -\frac{\sqrt{3}}{4} \quad \text{and} \quad \frac{\pi}{2} < \theta < \pi \][/tex]
### Part (a): Finding [tex]\(\sin(2\theta)\)[/tex]
The double-angle identity for sine is:
[tex]\[ \sin(2\theta) = 2 \sin(\theta) \cos(\theta) \][/tex]
First, we need to find [tex]\(\sin(\theta)\)[/tex]. Using the Pythagorean identity:
[tex]\[ \sin^2(\theta) + \cos^2(\theta) = 1 \][/tex]
Substitute [tex]\(\cos(\theta)\)[/tex] into the equation:
[tex]\[ \sin^2(\theta) + \left(-\frac{\sqrt{3}}{4}\right)^2 = 1 \][/tex]
[tex]\[ \sin^2(\theta) + \frac{3}{16} = 1 \][/tex]
[tex]\[ \sin^2(\theta) = 1 - \frac{3}{16} \][/tex]
[tex]\[ \sin^2(\theta) = \frac{16}{16} - \frac{3}{16} \][/tex]
[tex]\[ \sin^2(\theta) = \frac{13}{16} \][/tex]
Since [tex]\(\frac{\pi}{2} < \theta < \pi\)[/tex], [tex]\(\sin(\theta)\)[/tex] is positive:
[tex]\[ \sin(\theta) = \sqrt{\frac{13}{16}} = \frac{\sqrt{13}}{4} \][/tex]
Now, using the double-angle identity:
[tex]\[ \sin(2\theta) = 2 \sin(\theta) \cos(\theta) = 2 \left(\frac{\sqrt{13}}{4}\right) \left(-\frac{\sqrt{3}}{4}\right) \][/tex]
[tex]\[ \sin(2\theta) = 2 \left( -\frac{\sqrt{39}}{16} \right) = -\frac{\sqrt{39}}{8} \][/tex]
Thus, the exact value is approximately:
[tex]\[ \sin(2\theta) \approx -0.7806 \][/tex]
### Part (b): Finding [tex]\(\cos(2\theta)\)[/tex]
The double-angle identity for cosine is:
[tex]\[ \cos(2\theta) = 2 \cos^2(\theta) - 1 \][/tex]
Substitute [tex]\(\cos(\theta)\)[/tex]:
[tex]\[ \cos(2\theta) = 2 \left(-\frac{\sqrt{3}}{4}\right)^2 - 1 \][/tex]
[tex]\[ \cos(2\theta) = 2 \left(\frac{3}{16}\right) - 1 \][/tex]
[tex]\[ \cos(2\theta) = \frac{6}{16} - 1 \][/tex]
[tex]\[ \cos(2\theta) = \frac{6}{16} - \frac{16}{16} \][/tex]
[tex]\[ \cos(2\theta) = \frac{6 - 16}{16} \][/tex]
[tex]\[ \cos(2\theta) = \frac{-10}{16} \][/tex]
[tex]\[ \cos(2\theta) = -\frac{5}{8} \][/tex]
Thus, the exact value is:
[tex]\[ \cos(2\theta) = -0.625 \][/tex]
### Part (c): Finding [tex]\(\sin\left(\frac{\theta}{2}\right)\)[/tex]
The half-angle identity for sine is:
[tex]\[ \sin\left(\frac{\theta}{2}\right) = \sqrt{\frac{1 - \cos(\theta)}{2}} \][/tex]
Substitute [tex]\(\cos(\theta)\)[/tex]:
[tex]\[ \sin\left(\frac{\theta}{2}\right) = \sqrt{\frac{1 - \left(-\frac{\sqrt{3}}{4}\right)}{2}} \][/tex]
[tex]\[ \sin\left(\frac{\theta}{2}\right) = \sqrt{\frac{1 + \frac{\sqrt{3}}{4}}{2}} \][/tex]
[tex]\[ \sin\left(\frac{\theta}{2}\right) = \sqrt{\frac{\frac{4}{4} + \frac{\sqrt{3}}{4}}{2}} \][/tex]
[tex]\[ \sin\left(\frac{\theta}{2}\right) = \sqrt{\frac{\frac{4+\sqrt{3}}{4}}{2}} \][/tex]
[tex]\[ \sin\left(\frac{\theta}{2}\right) = \sqrt{\frac{4+\sqrt{3}}{8}} \][/tex]
[tex]\[ \sin\left(\frac{\theta}{2}\right) = \sqrt{\frac{4+\sqrt{3}}{8}} \][/tex]
Since [tex]\(\theta/2\)[/tex] is in the first quadrant, [tex]\(\sin(\theta/2)\)[/tex] is positive.
Thus, approximately:
[tex]\[ \sin\left(\frac{\theta}{2}\right) \approx 0.8465 \][/tex]
### Part (d): Finding [tex]\(\cos\left(\frac{\theta}{2}\right)\)[/tex]
The half-angle identity for cosine is:
[tex]\[ \cos\left(\frac{\theta}{2}\right) = \sqrt{\frac{1 + \cos(\theta)}{2}} \][/tex]
Substitute [tex]\(\cos(\theta)\)[/tex]:
[tex]\[ \cos\left(\frac{\theta}{2}\right) = \sqrt{\frac{1 + \left(-\frac{\sqrt{3}}{4}\right)}{2}} \][/tex]
[tex]\[ \cos\left(\frac{\theta}{2}\right) = \sqrt{\frac{1 - \frac{\sqrt{3}}{4}}{2}} \][/tex]
[tex]\[ \cos\left(\frac{\theta}{2}\right) = \sqrt{\frac{\frac{4}{4} - \frac{\sqrt{3}}{4}}{2}} \][/tex]
[tex]\[ \cos\left(\frac{\theta}{2}\right) = \sqrt{\frac{\frac{4-\sqrt{3}}{4}}{2}} \][/tex]
[tex]\[ \cos\left(\frac{\theta}{2}\right) = \sqrt{\frac{4-\sqrt{3}}{8}} \][/tex]
\]
\cos\left(\frac{\theta}{2}\right) = \sqrt{\frac{4-\sqrt{3}}{8}}
\]
Since [tex]\(\theta/2\)[/tex] is in the first quadrant, [tex]\(\cos(\theta/2)\)[/tex] is positive.
Thus, approximately:
[tex]\[ \cos\left(\frac{\theta}{2}\right) \approx 0.5324 \][/tex]
In summary:
[tex]\[ (a) \sin(2\theta) \approx -0.7806 \][/tex]
[tex]\[ (b) \cos(2\theta) = -0.625 \][/tex]
\]
(c) \sin\left(\frac{\theta}{2}\right) \approx 0.8465
\]
[tex]\[ (d) \cos\left(\frac{\theta}{2}\right) \approx 0.5324 \][/tex]
Given:
[tex]\[ \cos(\theta) = -\frac{\sqrt{3}}{4} \quad \text{and} \quad \frac{\pi}{2} < \theta < \pi \][/tex]
### Part (a): Finding [tex]\(\sin(2\theta)\)[/tex]
The double-angle identity for sine is:
[tex]\[ \sin(2\theta) = 2 \sin(\theta) \cos(\theta) \][/tex]
First, we need to find [tex]\(\sin(\theta)\)[/tex]. Using the Pythagorean identity:
[tex]\[ \sin^2(\theta) + \cos^2(\theta) = 1 \][/tex]
Substitute [tex]\(\cos(\theta)\)[/tex] into the equation:
[tex]\[ \sin^2(\theta) + \left(-\frac{\sqrt{3}}{4}\right)^2 = 1 \][/tex]
[tex]\[ \sin^2(\theta) + \frac{3}{16} = 1 \][/tex]
[tex]\[ \sin^2(\theta) = 1 - \frac{3}{16} \][/tex]
[tex]\[ \sin^2(\theta) = \frac{16}{16} - \frac{3}{16} \][/tex]
[tex]\[ \sin^2(\theta) = \frac{13}{16} \][/tex]
Since [tex]\(\frac{\pi}{2} < \theta < \pi\)[/tex], [tex]\(\sin(\theta)\)[/tex] is positive:
[tex]\[ \sin(\theta) = \sqrt{\frac{13}{16}} = \frac{\sqrt{13}}{4} \][/tex]
Now, using the double-angle identity:
[tex]\[ \sin(2\theta) = 2 \sin(\theta) \cos(\theta) = 2 \left(\frac{\sqrt{13}}{4}\right) \left(-\frac{\sqrt{3}}{4}\right) \][/tex]
[tex]\[ \sin(2\theta) = 2 \left( -\frac{\sqrt{39}}{16} \right) = -\frac{\sqrt{39}}{8} \][/tex]
Thus, the exact value is approximately:
[tex]\[ \sin(2\theta) \approx -0.7806 \][/tex]
### Part (b): Finding [tex]\(\cos(2\theta)\)[/tex]
The double-angle identity for cosine is:
[tex]\[ \cos(2\theta) = 2 \cos^2(\theta) - 1 \][/tex]
Substitute [tex]\(\cos(\theta)\)[/tex]:
[tex]\[ \cos(2\theta) = 2 \left(-\frac{\sqrt{3}}{4}\right)^2 - 1 \][/tex]
[tex]\[ \cos(2\theta) = 2 \left(\frac{3}{16}\right) - 1 \][/tex]
[tex]\[ \cos(2\theta) = \frac{6}{16} - 1 \][/tex]
[tex]\[ \cos(2\theta) = \frac{6}{16} - \frac{16}{16} \][/tex]
[tex]\[ \cos(2\theta) = \frac{6 - 16}{16} \][/tex]
[tex]\[ \cos(2\theta) = \frac{-10}{16} \][/tex]
[tex]\[ \cos(2\theta) = -\frac{5}{8} \][/tex]
Thus, the exact value is:
[tex]\[ \cos(2\theta) = -0.625 \][/tex]
### Part (c): Finding [tex]\(\sin\left(\frac{\theta}{2}\right)\)[/tex]
The half-angle identity for sine is:
[tex]\[ \sin\left(\frac{\theta}{2}\right) = \sqrt{\frac{1 - \cos(\theta)}{2}} \][/tex]
Substitute [tex]\(\cos(\theta)\)[/tex]:
[tex]\[ \sin\left(\frac{\theta}{2}\right) = \sqrt{\frac{1 - \left(-\frac{\sqrt{3}}{4}\right)}{2}} \][/tex]
[tex]\[ \sin\left(\frac{\theta}{2}\right) = \sqrt{\frac{1 + \frac{\sqrt{3}}{4}}{2}} \][/tex]
[tex]\[ \sin\left(\frac{\theta}{2}\right) = \sqrt{\frac{\frac{4}{4} + \frac{\sqrt{3}}{4}}{2}} \][/tex]
[tex]\[ \sin\left(\frac{\theta}{2}\right) = \sqrt{\frac{\frac{4+\sqrt{3}}{4}}{2}} \][/tex]
[tex]\[ \sin\left(\frac{\theta}{2}\right) = \sqrt{\frac{4+\sqrt{3}}{8}} \][/tex]
[tex]\[ \sin\left(\frac{\theta}{2}\right) = \sqrt{\frac{4+\sqrt{3}}{8}} \][/tex]
Since [tex]\(\theta/2\)[/tex] is in the first quadrant, [tex]\(\sin(\theta/2)\)[/tex] is positive.
Thus, approximately:
[tex]\[ \sin\left(\frac{\theta}{2}\right) \approx 0.8465 \][/tex]
### Part (d): Finding [tex]\(\cos\left(\frac{\theta}{2}\right)\)[/tex]
The half-angle identity for cosine is:
[tex]\[ \cos\left(\frac{\theta}{2}\right) = \sqrt{\frac{1 + \cos(\theta)}{2}} \][/tex]
Substitute [tex]\(\cos(\theta)\)[/tex]:
[tex]\[ \cos\left(\frac{\theta}{2}\right) = \sqrt{\frac{1 + \left(-\frac{\sqrt{3}}{4}\right)}{2}} \][/tex]
[tex]\[ \cos\left(\frac{\theta}{2}\right) = \sqrt{\frac{1 - \frac{\sqrt{3}}{4}}{2}} \][/tex]
[tex]\[ \cos\left(\frac{\theta}{2}\right) = \sqrt{\frac{\frac{4}{4} - \frac{\sqrt{3}}{4}}{2}} \][/tex]
[tex]\[ \cos\left(\frac{\theta}{2}\right) = \sqrt{\frac{\frac{4-\sqrt{3}}{4}}{2}} \][/tex]
[tex]\[ \cos\left(\frac{\theta}{2}\right) = \sqrt{\frac{4-\sqrt{3}}{8}} \][/tex]
\]
\cos\left(\frac{\theta}{2}\right) = \sqrt{\frac{4-\sqrt{3}}{8}}
\]
Since [tex]\(\theta/2\)[/tex] is in the first quadrant, [tex]\(\cos(\theta/2)\)[/tex] is positive.
Thus, approximately:
[tex]\[ \cos\left(\frac{\theta}{2}\right) \approx 0.5324 \][/tex]
In summary:
[tex]\[ (a) \sin(2\theta) \approx -0.7806 \][/tex]
[tex]\[ (b) \cos(2\theta) = -0.625 \][/tex]
\]
(c) \sin\left(\frac{\theta}{2}\right) \approx 0.8465
\]
[tex]\[ (d) \cos\left(\frac{\theta}{2}\right) \approx 0.5324 \][/tex]
Thank you for visiting. Our goal is to provide the most accurate answers for all your informational needs. Come back soon. Thanks for using our service. We're always here to provide accurate and up-to-date answers to all your queries. Westonci.ca is your trusted source for answers. Visit us again to find more information on diverse topics.
