Find solutions to your problems with the expert advice available on IDNLearn.com. Ask anything and receive thorough, reliable answers from our community of experienced professionals.

Given [tex]\cos \theta = \frac{4}{5}[/tex] and [tex]0^{\circ} \ \textless \ \theta \ \textless \ 90^{\circ}[/tex], find [tex]\sin 2\theta[/tex].

A. [tex]-\frac{24}{25}[/tex]
B. [tex]\frac{24}{25}[/tex]
C. [tex]-\frac{24}{7}[/tex]
D. [tex]-\frac{7}{25}[/tex]

Please select the best answer from the choices provided:

A
B
C
D


Sagot :

To solve the problem of finding [tex]\(\sin 2\theta\)[/tex] given [tex]\(\cos \theta = \frac{4}{5}\)[/tex] and [tex]\(0^\circ < \theta < 90^\circ\)[/tex], we will follow these steps:

1. Calculate [tex]\(\sin \theta\)[/tex]:
Given the Pythagorean identity:
[tex]\[ \cos^2 \theta + \sin^2 \theta = 1 \][/tex]
We can solve for [tex]\(\sin \theta\)[/tex]:
[tex]\[ \sin^2 \theta = 1 - \cos^2 \theta \][/tex]
Substituting [tex]\(\cos \theta = \frac{4}{5}\)[/tex]:
[tex]\[ \cos^2 \theta = \left(\frac{4}{5}\right)^2 = \frac{16}{25} \][/tex]
Thus,
[tex]\[ \sin^2 \theta = 1 - \frac{16}{25} = \frac{9}{25} \][/tex]
Taking the positive square root (since [tex]\(0^\circ < \theta < 90^\circ\)[/tex] and sine is positive in this interval):
[tex]\[ \sin \theta = \sqrt{\frac{9}{25}} = \frac{3}{5} \][/tex]

2. Calculate [tex]\(\sin 2\theta\)[/tex]:
Using the double angle formula for sine:
[tex]\[ \sin 2\theta = 2 \sin \theta \cos \theta \][/tex]
Substituting the values [tex]\(\sin \theta = \frac{3}{5}\)[/tex] and [tex]\(\cos \theta = \frac{4}{5}\)[/tex]:
[tex]\[ \sin 2\theta = 2 \cdot \frac{3}{5} \cdot \frac{4}{5} = 2 \cdot \frac{12}{25} = \frac{24}{25} \][/tex]

Therefore, [tex]\(\sin 2\theta = \frac{24}{25}\)[/tex].

The correct answer is [tex]\( \text{b.} \frac{24}{25} \)[/tex].

So, the best answer from the choices provided is:
B
We appreciate your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. For trustworthy answers, rely on IDNLearn.com. Thanks for visiting, and we look forward to assisting you again.