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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]
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