Question #a9735

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Question #a9735

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Answer 1 · 2016-03-05T15:17:51

$sin 2 θ = \frac{24}{25}, cos 2 θ = - \frac{7}{25}, tan 2 θ = - \frac{24}{7}$ $sin 2theta=24/25, cos 2theta=-7/25, tan 2theta=-24/7$

Explanation:

It is known that
$cos θ = \frac{3}{5}$ $cos theta =3/5$
And
$0 < θ < 90^{\circ}$ $0< theta<90^@$

We also know that
$sin θ = \sqrt{1 - {cos}^{2} θ}$ $sin theta = sqrt (1-cos^2 theta)$
Since $θ$ $theta$ is an angle of the first quadrant $sin θ > 0$ $sin theta>0$ and we have
$sin θ = (+) \sqrt{1 - {(\frac{3}{5})}^{2}} = \sqrt{1 - \frac{9}{25}} = \sqrt{\frac{16}{25}} = \frac{4}{5}$ $sin theta =(+)sqrt(1-(3/5)^2)=sqrt(1-9/25)=sqrt(16/25)=4/5$

It's also known that

$sin 2 θ = 2 sin θ \cdot cos θ$ $sin 2theta=2sin theta *cos theta$
=> $sin 2 θ = 2 \cdot \frac{4}{5} \cdot \frac{3}{5} = \frac{24}{25}$ $sin 2theta=2*4/5*3/5=24/25$
And that
$cos 2 θ = {cos}^{2} θ - {sin}^{2} θ = {(\frac{3}{5})}^{2} - {(\frac{4}{5})}^{2} = \frac{9 - 16}{25} = - \frac{7}{25}$ $cos 2theta=cos^2 theta-sin^2 theta=(3/5)^2-(4/5)^2=(9-16)/25=-7/25$

So

$tan 2theta=(sin 2theta)/(cos theta)=(24/cancel(25))/(-7/cancel(25))=-24/7$

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Question #a9735

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