 # find the exact values of \sin 2u, \cos 2u, and 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
You can still ask an expert for help

• Live experts 24/7
• Questions are typically answered in as fast as 30 minutes
• Personalized clear answers

Solve your problem for the price of one coffee

• Math expert for every subject
• Pay only if we can solve it poleglit3

Step 1
Given data are: $\mathrm{tan}u=\frac{3}{5}$, u is in 1. quadrant.
So we have:
$\mathrm{tan}u=\frac{3}{5}$
$\frac{\mathrm{sin}u}{\mathrm{cos}u}=\frac{3}{5}$
$⇒\mathrm{sin}u=\frac{3}{5}\cdot \mathrm{cos}u$
$⇒{\mathrm{sin}}^{2}u+{\mathrm{cos}}^{2}u=1$ use: $\mathrm{sin}u=\frac{3}{5}\cdot \mathrm{cos}u$
$\frac{34}{25}{\mathrm{cos}}^{2}u=1$
${\mathrm{cos}}^{2}u=\frac{\frac{25}{34}}{\sqrt{\left\{}}$
$\mathrm{cos}u=±\frac{\frac{5}{\sqrt{34}}}{u}$ is in 1. quadrant
$\mathrm{cos}u=\frac{5}{\sqrt{34}}⇒\mathrm{sin}u=\frac{3}{\sqrt{34}}$
Step 2
We can now calculate what is required in the:
$\mathrm{sin}2u=2\mathrm{sin}u\mathrm{cos}u$
$=2\cdot \frac{3}{\sqrt{34}}\cdot \frac{5}{\sqrt{34}}$
$=\frac{15}{17}$
$\mathrm{cos}2u={\mathrm{cos}}^{2}u-{\mathrm{sin}}^{2}u$
$=\frac{25}{34}-\frac{9}{34}$
$=\frac{8}{17}$
$\mathrm{tan}2u=\frac{2\mathrm{tan}u}{1-{\mathrm{tan}}^{2}u}$
$=\frac{2\cdot \frac{3}{5}}{1-\frac{9}{25}}$
$=\frac{15}{8}$

###### Not exactly what you’re looking for? censoratojk
Given: $\mathrm{tan}\left(u\right)=\frac{3}{5},0
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 $\sqrt{34}$.
$\mathrm{sin}\left(2u\right)=2\mathrm{sin}\left(u\right)\mathrm{cos}\left(u\right)=2\left(\frac{3}{\sqrt{34}}\right)\left(\frac{5}{\sqrt{34}}\right)=\frac{30}{34}=\frac{15}{17}$
$\mathrm{cos}\left(2u\right)=1-2{\mathrm{sin}}^{2}\left(u\right)=1-2{\left(\frac{3}{\sqrt{34}}\right)}^{2}=1-\frac{18}{34}=\frac{34}{34}-\frac{18}{34}=\frac{16}{34}=\frac{8}{17}$
$\mathrm{tan}\left(2u\right)=\frac{3\mathrm{tan}\left(u\right)}{1-{\mathrm{tan}}^{2}\left(u\right)}=\frac{2\left(\frac{3}{5}\right)}{1-{\left(\frac{3}{5}\right)}^{2}}=\frac{\frac{6}{5}}{1-\frac{9}{25}}=\frac{\frac{6}{5}}{\frac{25}{25}-\frac{9}{25}}=\frac{\frac{6}{5}}{\frac{16}{25}}=\frac{15}{8}$