Question #b3353

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Question #b3353

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Answer 1 · 2016-02-29T04:05:47

tan 2u = 240/161.

Explanation:

Think of a right triangle with sides 8, 15, 17 --

$8^{2} + 15^{2} = 17^{2}$ $8^{2}+15^{2} =17^{2}$ .

The angle at the hypoteneuse and the longer leg has a cosine of 15/17 and a sine of 8/17. Call that angle w.

Adding $π$ $\pi$ to w gives angle u between $π$ $\pi$ and $3 \frac{π}{2}$ $3\pi/2$ , with cos u = -cos w = -15/17. Then sin u = -sin w = -8/17. Dividing the sine by the cosine gives tan u = 8/15.

Now use the double angle formula,

tan 2u
= 2tan u/(1- ${tan}^{2} u$ $tan^{2} u$ )

We have tan u = 8/15. So tan 2u = (16/15)/(1-64/225) = 240/161.

Answer 2 · 2016-02-29T05:31:09

$tan 2 u = \frac{240}{161}$ $tan 2u = 240/161$

Explanation:

$cos u = - \frac{15}{17}$ $cos u = -15/17$
${sin}^{2} u = 1 - \frac{225}{289} = \pm \frac{64}{17}$ $sin^2 u = 1 - 225/289 = +- 64/17$
Since $π < u < \frac{3 π}{2}$ $pi < u < (3pi)/2$ , then $sin u = - \frac{8}{17}$ $sin u = -8/17$
$tan u = \frac{sin u}{cos u} = (- \frac{8}{17}) (- \frac{17}{15}) = \frac{8}{15}$ $tan u = (sin u)/(cos u) = (-8/17)(-17/15) = 8/15$
To find tan 2u apply the trig identity:
$tan 2 u = \frac{2 tan u}{1 - {tan}^{2} u}$ $tan 2u = (2tan u)/(1 - tan^2 u)$
Proceed numerically:
$2 tan u = 2 (\frac{8}{15}) = \frac{16}{15}$ $2tan u = 2(8/15) = 16/15$
$(1 - {tan}^{2} u) = 1 - \frac{64}{225} = \frac{161}{225}$ $(1 - tan^2 u) = 1 - 64/225 = 161/225$
Answer: $tan 2 u = (\frac{16}{15}) (\frac{225}{161}) = \frac{240}{161}$ $tan 2u = (16/15)(225/161) = 240/161$

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Question #b3353

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