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To find [tex]\(\sin 2\theta\)[/tex] given that [tex]\(\tan \theta = \frac{11}{6}\)[/tex] and that [tex]\(\theta\)[/tex] is in the first quadrant, we follow these steps:
1. Find [tex]\(\sin \theta\)[/tex] and [tex]\(\cos \theta\)[/tex]:
Since [tex]\(\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{11}{6}\)[/tex], we can use the identity [tex]\(1 + \tan^2 \theta = \frac{1}{\cos^2 \theta}\)[/tex] to help find [tex]\(\sin \theta\)[/tex] and [tex]\(\cos \theta\)[/tex].
2. Calculate [tex]\(\sin \theta\)[/tex] and [tex]\(\cos \theta\)[/tex]:
Given [tex]\(\tan \theta = \frac{11}{6}\)[/tex],
[tex]\[ \tan^2 \theta = \left(\frac{11}{6}\right)^2 = \frac{121}{36} \][/tex]
Using the Pythagorean identity, we have:
[tex]\[ 1 + \tan^2 \theta = \frac{1}{\cos^2 \theta} \][/tex]
Thus,
[tex]\[ 1 + \frac{121}{36} = \frac{1}{\cos^2 \theta} \][/tex]
So,
[tex]\[ \frac{36 + 121}{36} = \frac{1}{\cos^2 \theta} \][/tex]
[tex]\[ \frac{157}{36} = \frac{1}{\cos^2 \theta} \][/tex]
Hence,
[tex]\[ \cos^2 \theta = \frac{36}{157} \][/tex]
and therefore,
[tex]\[ \cos \theta = \sqrt{\frac{36}{157}} = \frac{6}{\sqrt{157}} \][/tex]
Using [tex]\(\sin^2 \theta + \cos^2 \theta = 1\)[/tex],
[tex]\[ \sin^2 \theta = 1 - \cos^2 \theta = 1 - \frac{36}{157} = \frac{157 - 36}{157} = \frac{121}{157} \][/tex]
Thus,
[tex]\[ \sin \theta = \sqrt{\frac{121}{157}} = \frac{11}{\sqrt{157}} \][/tex]
3. Compute [tex]\(\sin 2\theta\)[/tex] using the double-angle formula:
The double-angle formula for sine is:
[tex]\[ \sin 2\theta = 2 \sin \theta \cos \theta \][/tex]
Substituting the found values:
[tex]\[ \sin \theta = \frac{11}{\sqrt{157}} \][/tex]
and
[tex]\[ \cos \theta = \frac{6}{\sqrt{157}} \][/tex]
Therefore,
[tex]\[ \sin 2\theta = 2 \left(\frac{11}{\sqrt{157}}\right) \left(\frac{6}{\sqrt{157}}\right) = 2 \left(\frac{66}{157}\right) = \frac{132}{157} \][/tex]
So, [tex]\(\sin 2\theta = 0.8407643312101911\)[/tex] (approximately).
Thus, the value of [tex]\(\sin 2\theta\)[/tex] is [tex]\(\boxed{0.8407643312101911}\)[/tex].
1. Find [tex]\(\sin \theta\)[/tex] and [tex]\(\cos \theta\)[/tex]:
Since [tex]\(\tan \theta = \frac{\sin \theta}{\cos \theta} = \frac{11}{6}\)[/tex], we can use the identity [tex]\(1 + \tan^2 \theta = \frac{1}{\cos^2 \theta}\)[/tex] to help find [tex]\(\sin \theta\)[/tex] and [tex]\(\cos \theta\)[/tex].
2. Calculate [tex]\(\sin \theta\)[/tex] and [tex]\(\cos \theta\)[/tex]:
Given [tex]\(\tan \theta = \frac{11}{6}\)[/tex],
[tex]\[ \tan^2 \theta = \left(\frac{11}{6}\right)^2 = \frac{121}{36} \][/tex]
Using the Pythagorean identity, we have:
[tex]\[ 1 + \tan^2 \theta = \frac{1}{\cos^2 \theta} \][/tex]
Thus,
[tex]\[ 1 + \frac{121}{36} = \frac{1}{\cos^2 \theta} \][/tex]
So,
[tex]\[ \frac{36 + 121}{36} = \frac{1}{\cos^2 \theta} \][/tex]
[tex]\[ \frac{157}{36} = \frac{1}{\cos^2 \theta} \][/tex]
Hence,
[tex]\[ \cos^2 \theta = \frac{36}{157} \][/tex]
and therefore,
[tex]\[ \cos \theta = \sqrt{\frac{36}{157}} = \frac{6}{\sqrt{157}} \][/tex]
Using [tex]\(\sin^2 \theta + \cos^2 \theta = 1\)[/tex],
[tex]\[ \sin^2 \theta = 1 - \cos^2 \theta = 1 - \frac{36}{157} = \frac{157 - 36}{157} = \frac{121}{157} \][/tex]
Thus,
[tex]\[ \sin \theta = \sqrt{\frac{121}{157}} = \frac{11}{\sqrt{157}} \][/tex]
3. Compute [tex]\(\sin 2\theta\)[/tex] using the double-angle formula:
The double-angle formula for sine is:
[tex]\[ \sin 2\theta = 2 \sin \theta \cos \theta \][/tex]
Substituting the found values:
[tex]\[ \sin \theta = \frac{11}{\sqrt{157}} \][/tex]
and
[tex]\[ \cos \theta = \frac{6}{\sqrt{157}} \][/tex]
Therefore,
[tex]\[ \sin 2\theta = 2 \left(\frac{11}{\sqrt{157}}\right) \left(\frac{6}{\sqrt{157}}\right) = 2 \left(\frac{66}{157}\right) = \frac{132}{157} \][/tex]
So, [tex]\(\sin 2\theta = 0.8407643312101911\)[/tex] (approximately).
Thus, the value of [tex]\(\sin 2\theta\)[/tex] is [tex]\(\boxed{0.8407643312101911}\)[/tex].
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