Question

An angle A in standard position has a terminal side which passes through the point (-6,-3). Determine values for the following: sin(A) = cos(A) =

Answer 1

Given an angle θ in standard position passing through (x, y), the following relationship stands:

$\tan\theta=\frac{y}{x}$

Our angle has a terminal side that passes through (-6, -3). Since both x and y are negative, the angle lies in quadrant III where the sine and the cosine are negative.

Substituting the given values:

$\tan\theta=\frac{-3}{-6}=\frac{1}{2}$

Now we use the following identity:

$\sec^2\theta=1+\tan^2\theta$

Substituting:

$\begin{gathered} \sec^2\theta=1+(\frac{1}{2})^2 \\ \text{ Calculating:} \\ \sec^2\theta=1+\frac{1}{4}=\frac{5}{4} \end{gathered}$

The secant is also negative in quadrant III, so:

$\begin{gathered} \sec\theta=-\sqrt{\frac{5}{4}} \\ \sec\theta=-\frac{\sqrt{5}}{2} \end{gathered}$

The cosine is the reciprocal of the secant:

$\cos\theta=\frac{1}{\sec\theta}=\frac{1}{(-\frac{\sqrt{5}}{2})}=-\frac{2}{\sqrt{5}}$

Rationalizing:

$\boxed{\cos\theta={}-\frac{2\sqrt{5}}{5}}$

The sine can be calculated as:

$\begin{gathered} \sin\theta=\tan\theta\cdot\cos\theta \\ \sin\theta=\frac{1}{2}\cdot(-\frac{2\sqrt{5}}{5}) \\ \sin\theta=-\frac{\sqrt{5}}{5} \end{gathered}$$\boxed{\sin\theta=-\frac{\sqrt{5}}{5}}$