# Given tanx is 3/4. you can use the double formula to find tan2x. the teacher wants us to find 3 other doubles like sin2x or cos2x without using the double angle formulas.

Given tanx is $\frac{3}{4}$. you can use the double formula to find $\mathrm{tan}2x$. the teacher wants us to find 3 other doubles like $\mathrm{sin}2x$ or $\mathrm{cos}2x$ without using the double angle formulas.
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Latisha Oneil
If $\mathrm{tan}\left(x\right)=\frac{3}{4}$, then $\mathrm{sin}\left(x\right)=\frac{3}{5}\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}\mathrm{cos}\left(x\right)=\frac{4}{5}$, because of Pythagoras’ Theorem.
This is the 3-4-5 right triangle. $\mathrm{tan}\left(2x\right)=2\frac{\mathrm{tan}\left(x\right)}{1-{\mathrm{tan}}^{2}\left(x\right)}=\frac{\frac{3}{2}}{1-\frac{9}{16}}=\frac{24}{7}$.
If the legs of a right triangle are 24 and 7, the hypotenuse is $\sqrt{{24}^{2}+{7}^{2}}=\sqrt{576+49}=\sqrt{625}=25$.
So $\mathrm{sin}\left(2x\right)=\frac{24}{25}$ and $\mathrm{cos}\left(2x\right)=\frac{7}{25}.$
$\mathrm{cot}\left(2x\right)=\frac{7}{24},\mathrm{csc}\left(2x\right)=\frac{25}{24},\mathrm{sec}\left(2x\right)=\frac{25}{7}.$