Question

Suppose cos(x)= -1/3, where π/2 ≤ x ≤ π. What is the value of tan(2x). EDGE

Answer 1

Answer:

D

Step-by-step explanation:

We are given that:

$\displaystyle \cos x = -\frac{1}{3}\text{ where } \pi /2 \leq x \leq \pi$

And we want to find the value of tan(2x).

Note that since x is between π/2 and π, it is in QII.

In QII, cosine and tangent are negative and only sine is positive.

We can rewrite our expression as:

$\displaystyle \tan(2x)=\frac{\sin(2x)}{\cos(2x)}$

Using double angle identities:

$\displaystyle \tan(2x)=\frac{2\sin x\cos x}{\cos^2 x-\sin^2 x}$

Since cosine relates the ratio of the adjacent side to the hypotenuse and we are given that cos(x) = -1/3, this means that our adjacent side is one and our hypotenuse is three (we can ignore the negative). Using this information, find the opposite side:

$\displaystyle o=\sqrt{3^2-1^2}=\sqrt{8}=2\sqrt{2}$

So, our adjacent side is 1, our opposite side is 2√2, and our hypotenuse is 3.

From the above information, substitute in appropriate values. And since x is in QII, cosine and tangent will be negative while sine will be positive. Hence:

$\displaystyle \tan(2x)=\frac{2(2\sqrt{2}/3)(-1/3)}{(-1/3)^2-(2\sqrt{2}/3)^2}$

Simplify:

$\displaystyle \tan(2x)=\frac{-4\sqrt{2}/9}{(1/9)-(8/9)}$

Evaluate:

$\displaystyle \tan(2x)=\frac{-4\sqrt{2}/9}{-7/9} = \frac{4\sqrt{2}}{7}$

The final answer is positive, so we can eliminate A and B.

We can simplify D to:

$\displaystyle \frac{2\sqrt{8}}{7}=\frac{2(2\sqrt{2}}{7}=\frac{4\sqrt{2}}{7}$

So, our answer is D.