Question

Find sin 2x, cos 2x, and tan 2x from the given information. tan x = − 1 4 , cos x > 0

Answer 1

so what's the question?

Answer 2

lol im confused

Answer 3

@Saisuke<3

Answer 4

If \(\tan x < 0\) and \(\cos x>0\), that means the angle \(x\) is in Quadrant IV. (Therefore, only \(\cos x,\sec x>0\); all the other trig functions will be negative valued).

We'll find \(\sin(2x)\) and \(\cos(2x)\) first; To do that, though, we need to know \(\cos x\) and \(\sin x\).

Recall that \(\tan^2 x+ 1 = \sec^2 x\). Thus, \(\sec^2 x = (-\frac{1}{4})^2 +1 = \frac{17}{16}\implies \cos^2 x = \frac{16}{17}\). Therefore \(\cos x = \sqrt{\dfrac{16}{17}} = \dfrac{4}{\sqrt{17}}\). Now, we recall that \(\tan x = \dfrac{\sin x}{\cos x}\). Hence
\(\sin x = \tan x \cos x = -\dfrac{1}{4}\cdot\dfrac{4}{\sqrt{17}} = -\dfrac{1}{\sqrt{17}}\).

Next, we note that

\(\sin(2x)=2\sin x\cos x = 2\cdot\dfrac{-1}{\sqrt{17}}\cdot \dfrac{4}{\sqrt{17}} = -\dfrac{4}{17} \)

and

\(\cos(2x) = \cos^2 x - \sin^2 x =\dfrac{16}{17} - \dfrac{1}{17} = \dfrac{15}{17}\)

Last but not least, we now have that \(\tan(2x) = \dfrac{\sin(2x)}{\cos(2x)} = \dfrac{-4/17}{15/17} =-\dfrac{4}{15}\).

Does this make sense? :-)

Answer 5

Woops; \(\sin(2x) = -\dfrac{\color{red}{8}}{17}\) and as a result \(\tan(2x) = -\dfrac{\color{red}{8}}{15}\). Sorry about that!