Question

i need the answer to this problem for my son so I can quiz him.

Answer 1

We know that

$\cot x=\frac{\cos x}{\sin x}$

Since cot x is negative, and we're given that cos x is postive, we have no alternative but for sin x to be negative

$-\cot x=\frac{\cos x}{-\sin x}$

Let's keep this in mind. Now, let's solve the equation we're given:

$\cot x=-\frac{\sqrt[]{3}}{2}\rightarrow\frac{\cos x}{\sin x}=-\frac{\sqrt[]{3}}{2}$

Let's square both sides, so we get

$\frac{\cos x}{\sin x}=-\frac{\sqrt[]{3}}{2}\Rightarrow\frac{\cos^2x}{\sin^2x}=\frac{3}{4}$

Now, we can use the Pythagorean identity that tells us that

$\cos ^2x=1-\sin ^2x$

This way, we'll have that

$\frac{\cos^2x}{\sin^2x}=\frac{3}{4}\rightarrow\frac{1-\sin^2x}{\sin^2x}=\frac{3}{4}$

Solving for sin x :

$\begin{gathered} \frac{1-\sin^2x}{\sin^2x}=\frac{3}{4}\rightarrow4(1-\sin ^2x)=3\sin ^2x \\ \rightarrow4-4\sin ^2x=3\sin ^2x\rightarrow4=7\sin ^2x\rightarrow\frac{4}{7}=\sin ^2x \\ \end{gathered}$

Taking square root in both sides, we get:

$\sin x=\pm\frac{2}{\sqrt[]{7}}\rightarrow\sin x=\pm\frac{2\text{ }\sqrt[]{7}}{7}$

Notice that we've stablished from the begining that sin x is negative. Therefore, we can conclude that:

$\sin x=-\frac{2\text{ }\sqrt[]{7}}{7}$