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# find teh exact values of sin2u, cos2u, and tan 2u --> using double angle formuals

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sin u = -4/5 , pi < u < 3pi/2

sin2u = 24/25

cos2u = -17/25

tan 2u = 4sqrt21/ 17

how to do??

Guest May 17, 2017
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#1
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sin u = -4/5 , pi < u < 3pi/2

sin2u = 24/25

cos2u = -17/25  not correct

tan 2u = 4sqrt21/ 17

how to do?

$$sin \ 2u=2sin \ u \cdot cos \ u\\cos \ u=\sqrt{1-sin^2u}\\sin\ 2u=2sin\ u\cdot\sqrt{1-sin^2u}$$

$$sin \ 2u=-\frac{8}{5}\cdot\sqrt{1-\frac{16}{25}}=-\frac{8}{5}\cdot -\sqrt{\frac{9}{25}}=-\frac{8}{5}\cdot -\frac{3}{5}$$

$$sin\ 2u=\frac{24}{25}$$

$$cos \ 2u=1-2sin^2u=1-2\cdot\frac{16}{25}$$

$$cos\ 2u=-\frac{7}{25}$$

$$tan \ 2u=\ \frac{2\cdot tan\ u}{1-tan^2u}\\tan\ u=\pm\ \frac{sin\ u}{\sqrt{1-sin^2u}}$$

$$tan^2u=\frac{sin^2u}{1-sin^2u}=\frac{\frac{16}{25}}{1-\frac{16}{25}}=\frac{16\cdot 25}{25\cdot 9}$$

$$tan\ u =\frac{4}{3}$$

$$tan\ 2u=\frac{\frac{8}{3}}{1-\frac{16}{9}}=-\ \frac{8\cdot 9 }{3\cdot 7}$$

$$tan\ 2u=-\ \frac{24}{7}$$

!

asinus  May 17, 2017

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