How do you find the values of $sin 2 θ$ $sin 2theta$ and $cos 2 θ$ $cos 2theta$ when $cos θ = \frac{12}{13}$ $cos theta = 12/13$ ?

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How do you find the values of $sin 2 θ$ $sin 2theta$ and $cos 2 θ$ $cos 2theta$ when $cos θ = \frac{12}{13}$ $cos theta = 12/13$ ?

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Answer 1 · 2018-08-04T00:17:44

Below

Explanation:

$θ$ $theta$ can be in the first quadrant $0 \leq θ \leq 90$ $0<=theta<=90$ or the fourth quadrant $270 \leq θ \leq 360$ $270<=theta<=360$

If $θ$ $theta$ is in the first quadrant,
then
$sin θ = \frac{5}{13}$ $sintheta=5/13$
$cos θ = \frac{12}{13}$ $costheta=12/13$
$tan θ = \frac{5}{12}$ $tantheta=5/12$

Therefore,
$sin 2 θ = 2 sin θ cos θ = 2 \times \frac{5}{13} \times \frac{12}{13} = \frac{120}{169}$ $sin2theta=2sinthetacostheta=2times5/13times12/13=120/169$

$cos 2 θ = {cos}^{2} θ - {sin}^{2} θ = {(\frac{12}{13})}^{2} - {(\frac{5}{13})}^{2} = \frac{144}{169} - \frac{25}{169} = \frac{119}{169}$ $cos2theta=cos^2theta-sin^2theta=(12/13)^2-(5/13)^2=144/169-25/169=119/169$

If $θ$ $theta$ is in the fourth quadrant,
then
$sin θ = - \frac{5}{13}$ $sintheta=-5/13$
$cos θ = \frac{12}{13}$ $costheta=12/13$
$tan θ = - \frac{5}{12}$ $tantheta=-5/12$

Therefore,
$sin 2 θ = 2 sin θ cos θ = 2 \times - \frac{5}{13} \times \frac{12}{13} = - \frac{120}{169}$ $sin2theta=2sinthetacostheta=2times-5/13times12/13=-120/169$

$cos 2 θ = {cos}^{2} θ - {sin}^{2} θ = {(\frac{12}{13})}^{2} - {(- \frac{5}{13})}^{2} = \frac{144}{169} - \frac{25}{169} = \frac{119}{169}$ $cos2theta=cos^2theta-sin^2theta=(12/13)^2-(-5/13)^2=144/169-25/169=119/169$

Answer 2 · 2018-08-04T02:12:03

$sin 2 θ = \frac{120}{169}, θ \in Q_{1} and - \frac{120}{169}, θ \in Q_{4}$ $sin 2theta = 120/169, theta in Q_1 and - 120/169, theta in Q_4$ .

$cos 2 θ = \frac{119}{169}$ $cos 2theta = 119/169$

Explanation:

See my answer in

https://socratic.org/questions/if-cos-a-5-13-how-do-you-find-sina-and-tana

As a continuation,

$sin θ = \frac{5}{13}, θ \in Q_{1}$ $sin theta = 5/13, theta in Q_1$ and it is $- \frac{5}{13}, θ \in Q_{4}$ $- 5/13, theta in Q_4$ .

$sin 2 θ = 2 sin θ cos θ = 2 (\frac{5}{13}) (\frac{12}{13})$ $sin 2theta = 2 sin theta cos theta = 2 ( 5/13)(12/13)$

$= \frac{120}{169}, θ \in Q_{1} and$ $= 120/169, theta in Q_1 and$

$= 2 (- \frac{5}{13}) (\frac{12}{13}) = - \frac{120}{169}, θ \in Q_{4}$ $= 2 ( -5/13)(12/13) = - 120/169, theta in Q_4$ .

$cos 2 θ = {cos}^{2} θ - {sin}^{2} θ = {(\frac{12}{13})}^{2} - {(\pm \frac{5}{13})}^{2}$ $cos 2theta = cos^2theta - sin^2theta = (12/13)^2 - (+-5/13)^2$

$= \frac{119}{169}$ $= 119/169$

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How do you find the values of $sin 2 θ$ $sin 2theta$ and $cos 2 θ$ $cos 2theta$ when $cos θ = \frac{12}{13}$ $cos theta = 12/13$ ?

2 Answers

Explanation:

Explanation:

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How do you find the values of sin2θsin 2theta and cos2θcos 2theta when cosθ=1213cos theta = 12/13?

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How do you find the values of $sin 2 θ$ $sin 2theta$ and $cos 2 θ$ $cos 2theta$ when $cos θ = \frac{12}{13}$ $cos theta = 12/13$ ?