Algotssleeddynf
2021-12-15
Answered

find the exact values of $\mathrm{sin}2u,\mathrm{cos}2u$ , and $\mathrm{tan}2u$ using the double-angle formulas. $\mathrm{tan}u=\frac{3}{5},0<u<\frac{\pi}{2}$

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poleglit3

Answered 2021-12-16
Author has **32** answers

Step 1

Given data are:

So we have:

Step 2

We can now calculate what is required in the:

censoratojk

Answered 2021-12-17
Author has **46** answers

Given: $\mathrm{tan}\left(u\right)=\frac{3}{5},0<u<\frac{\pi}{2}$

Angle u is in quadrant 1. A right triangle can be drawn is quadrant 1. Since$\mathrm{tan}\left(u\right)=\frac{3}{5}$ , the side opposite angle u is 3 and the side adjacent to angle u is 5. Using the pythagorean theorem the hypotenuse of the triangle is $\sqrt{34}$ .

$\mathrm{sin}\left(2u\right)=2\mathrm{sin}\left(u\right)\mathrm{cos}\left(u\right)=2\left(\frac{3}{\sqrt{34}}\right)\left(\frac{5}{\sqrt{34}}\right)=\frac{30}{34}=\frac{15}{17}$

$\mathrm{cos}\left(2u\right)=1-2{\mathrm{sin}}^{2}\left(u\right)=1-2{\left(\frac{3}{\sqrt{34}}\right)}^{2}=1-\frac{18}{34}=\frac{34}{34}-\frac{18}{34}=\frac{16}{34}=\frac{8}{17}$

$\mathrm{tan}\left(2u\right)=\frac{3\mathrm{tan}\left(u\right)}{1-{\mathrm{tan}}^{2}\left(u\right)}=\frac{2\left(\frac{3}{5}\right)}{1-{\left(\frac{3}{5}\right)}^{2}}=\frac{\frac{6}{5}}{1-\frac{9}{25}}=\frac{\frac{6}{5}}{\frac{25}{25}-\frac{9}{25}}=\frac{\frac{6}{5}}{\frac{16}{25}}=\frac{15}{8}$

Angle u is in quadrant 1. A right triangle can be drawn is quadrant 1. Since

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