Respuesta :

WojtekR
First calculate [tex]\sin\theta[/tex].

[tex]\sin^2\theta+\cos^2\theta=1\\\\\sin^2\theta=1-\cos^2\theta\\\\\sin^2\theta=1-\left(\dfrac{4}{5}\right)^2\\\\\\\sin^2\theta=1-\dfrac{16}{25}\\\\\\\sin^2\theta=\dfrac{9}{25}\quad|\sqrt{(\dots)}\\\\\\\sin\theta=\sqrt{\dfrac{9}{25}}\\\\\\\boxed{\sin\theta=\dfrac{3}{5}}[/tex]

We take positive value of sinθ because 0° < θ < 90°
Now we could calculate sin2θ, cos2θ and tan2θ:

[tex]\sin2\theta=2\sin\theta\cos\theta=2\cdot\dfrac{3}{5}\cdot\dfrac{4}{5}=\dfrac{2\cdot3\cdot4}{5\cdot5}=\dfrac{24}{25}\\\\\\\cos2\theta=\cos^2\theta-\sin^2\theta=\left(\dfrac{4}{5}\right)^2-\left(\dfrac{3}{5}\right)^2=\dfrac{16}{25}-\dfrac{9}{25}=\dfrac{7}{25}\\\\\\ \tan2\theta=\dfrac{\sin2\theta}{\cos2\theta}=\dfrac{\frac{24}{25}}{\frac{7}{25}}=\dfrac{24\cdot25}{7\cdot25}=\dfrac{24}{7}=3\dfrac{3}{7}[/tex]

Answer:

D edge

Step-by-step explanation:

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