# Thread: Find sin2x for each given the following condition:

1. ## Find sin2x for each given the following condition:

Find sin2x for each given the following condition:

a) sin x = 4/5 for 90o < x < 180o

b)tan x = -2 for 3pi/2 < x < 2pi

3. ## Re: Find sin2x for each given the following condition:

Hello, alejandro!

$\displaystyle \text{Find }\sin2x\text{ for each given the following conditions:}$

$\displaystyle a)\;\sin x \,=\,\tfrac{4}{5}\:\text{ for }90^o < x < 180^o$

We are given: .$\displaystyle \sin x \:=\:\tfrac{4}{5} \:=\:\tfrac{opp}{hyp}$

Angle $\displaystyle x$ is in a right triangle with: $\displaystyle opp = 3,\;hyp = 5.$
Pythagorus says: $\displaystyle adj = \pm4 \quad\Rightarrow\quad \cos x \,=\,\pm\tfrac{4}{5}$
Since $\displaystyle x$ in Quadrant 2, $\displaystyle \cos x \,=\,-\tfrac{4}{5}$

Identity: .$\displaystyle \sin2x \,=\,2\sin x\cos x$

Therefore: .$\displaystyle \sin2x \;=\;2\left(\frac{3}{5}\right)\left(-\frac{4}{5}\right) \;=\; -\frac{24}{25}$

$\displaystyle b)\;\tan x \,=\, -2\:\text{ for }\tfrac{3\pi}{2} < x < 2\pi$

Since $\displaystyle x$ is in Quadrant 4, we have: .$\displaystyle \tan x \,=\,\tfrac{\text{-}2}{1} \,=\,\tfrac{opp}{adj}$

Angle $\displaystyle x$ is in a right triangle with: $\displaystyle opp = \text{-}2,\;adj = 1$
Pythagorus says: .$\displaystyle hyp \,=\,\sqrt{5} \quad\Rightarrow\quad \sin x \,=\,\text{-}\tfrac{2}{\sqrt{5}},\;\cos x \,=\,\tfrac{1}{\sqrt{5}}$

Therefore: .$\displaystyle \sin 2x \:=\:2\left(-\frac{2}{\sqrt{5}}\right)\left(\frac{1}{\sqrt{5}} \right) \;=\; -\frac{4}{5}$