Question

Let 0 be an angle in quadrant IV such that cot 0 = -2/7. Find the exact values of sin 0 and sec 0

Answer 1

In quadrant IV, cosine is positive and sine is negative. This means that

$\cos^2(x)+\sin^2(x)=1 \iff \sin^2(x)=1-\cos^2(x) \implies \sin(x)=-\sqrt{1-\cos^2(x)$

The cotangent is defined as the ratio between the cosine and sine:

$\cot(x)=\dfrac{\cos(x)}{\sin(x)}=\dfrac{\cos(x)}{-\sqrt{1-\cos^2(x)}}=-\dfrac{2}{7}\iff \dfrac{\cos(x)}{\sqrt{1-\cos^2(x)}}=\dfrac{2}{7}$

So, we have the following equation:

$2\sqrt{1-\cos^2(x)}=7\cos(x)$

Squaring both sides yields

$4(1-\cos^2(x))=49\cos^2(x) \iff 4-4\cos^2(x)=49\cos^2(x)\iff 53\cos^2(x)=4$

The solution to this equation would be

$\cos^2(x)=\dfrac{4}{53}\iff \cos^2(x)=\pm\dfrac{2}{\sqrt{53}}$

But we know that the cosine has to be positive, so we have

$\cos(x)=\dfrac{2}{\sqrt{53}}$

And

$\sin(x)=-\sqrt{1-\dfrac{4}{53}}=-\sqrt{\dfrac{49}{53}}=-\dfrac{7}{\sqrt{53}}$

Finally, the secant is the inverse of the cosine, so it's

$\sec(x)=\dfrac{1}{\cos(x)}=\dfrac{\sqrt{53}}{2}$