Respuesta :

isyllus

Answer:

[tex]\tan2\theta=\pm\dfrac{4\sqrt{2}}{7}[/tex]

Step-by-step explanation:

Given: [tex]\sin\theta = \dfrac{1}{3}[/tex]

[tex]0\leq \theta \leq \dfrac{\pi}{2}[/tex]

Theta lie in first quadrant.

Multiply both sides by 2

[tex]0\leq 2\theta \leq \pi[/tex]

2theta lie in I and II quadrant.

[tex]\tan2\theta[/tex] is positive in I and negative in II quadrant.

[tex]\sin\theta=\dfrac{1}{3}[/tex]

[tex]\cos\theta=\dfrac{2\sqrt{2}}{3}[/tex]

[tex]\sin2\theta=2\sin\theta\cos\theta[/tex]

[tex]\sin2\theta=2\cdot \dfrac{1}{3}\cdot \dfrac{2\sqrt{2}}{3}=\dfrac{4\sqrt{2}}{9}[/tex]

[tex]\cos2\theta=\pm \sqrt{1-\sin^22\theta}=\pm\dfrac{7}{9}[/tex]

[tex]\tan2\theta=\dfrac{\sin2\theta}{\cos2\theta}[/tex]

[tex]\tan2\theta=\pm\dfrac{4\sqrt{2}}{7}[/tex]

Hence, The value of [tex]\tan2\theta=\pm\dfrac{4\sqrt{2}}{7}[/tex]

MrRoyal

The value of [tex]\tan(2\theta)[/tex] is [tex]\frac{4\sqrt 2}{7}[/tex]

The given parameter is:

[tex]\sin(\theta) = \frac 13[/tex]

Calculate cos(theta) using the following trigonometry ratio

[tex]\cos^2(\theta) = 1 - \sin^2(\theta)[/tex]

So, we have:

[tex]\cos^2(\theta) = 1 - (1/3)^2[/tex]

[tex]\cos^2(\theta) = 1 - \frac 19[/tex]

Subtract

[tex]\cos^2(\theta) = \frac 89[/tex]

Take the square roots of both sides

[tex]\cos(\theta) = \frac{2}{3}\sqrt 2[/tex]

tan (2 theta) is then calculated as:

[tex]\tan(2\theta) = \frac{\sin(2\theta)}{\cos(2\theta)}[/tex]

This gives

[tex]\tan(2\theta) = \frac{2\sin(\theta)\cos(\theta)}{\cos^2(\theta) - \sin^2(\theta)}[/tex]

So, we have:

[tex]\tan(2\theta) = \frac{2 * 1/3 * 2/3 \sqrt 2}{8/9 - 1/9}[/tex]

Simplify

[tex]\tan(2\theta) = \frac{4/9 \sqrt 2}{7/9}[/tex]

Further, simplify

[tex]\tan(2\theta) = \frac{4\sqrt 2}{7}[/tex]

Hence, the value of [tex]\tan(2\theta)[/tex] is [tex]\frac{4\sqrt 2}{7}[/tex]

Read more about trigonometry ratios at:

https://brainly.com/question/11967894

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