Question

Given cos(α)=7/9 and sin(α)<0, calculate the exact value of cot(α).

Answer 1

From the identity

$\sin ^2(\alpha)+\cos ^2(\alpha)=1$

we can find the value for sin(α). Using our value for the cosine and solving the identity for sin(α), we have

$\begin{gathered} \sin ^2(\alpha)+(\frac{7}{9})^2=1 \\ \sin ^2(\alpha)=1-(\frac{7}{9})^2 \\ \sin ^2(\alpha)=1-\frac{7^2}{9^2} \\ \sin ^2(\alpha)=1-\frac{49}{81} \\ \sin ^2(\alpha)=\frac{81}{81}-\frac{49}{81} \\ \sin ^2(\alpha)=\frac{32}{81} \\ \sin (\alpha)=\pm_{}\sqrt[]{\frac{32}{81}} \end{gathered}$

Since sin(α) < 0, we have

$\sin (\alpha)=-_{}\sqrt[]{\frac{32}{81}}=-\frac{4\sqrt[]{2}}{9}$

The cot(α) is defined as the ratio between the cosine and sine of alpha.

$\cot (\alpha)=\frac{\cos(\alpha)}{\sin(\alpha)}$

Then, the cot(α) is

$\cot (\alpha)=\frac{\frac{7}{9}}{-\frac{4\sqrt[]{2}}{9}}=-\frac{7}{4\sqrt[]{2}}=-\frac{7\sqrt[]{2}}{8}$

This is the exact value of cot(α).

$\cot (\alpha)=-\frac{7\sqrt[]{2}}{8}$