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=(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=(-3)/(-6)=(1)/(2)`

Now we use the following identity:

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

Substituting:

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

The secant is also negative in quadrant III, so:

`\begin{gathered} \sec\theta=-\sqrt{(5)/(4)} \\ \sec\theta=-(√(5))/(2) \end{gathered}`

The cosine is the reciprocal of the secant:

`\cos\theta=(1)/(\sec\theta)=(1)/((-(√(5))/(2)))=-(2)/(√(5))`

Rationalizing:

`\boxed{\cos\theta={}-(2√(5))/(5)}`

The sine can be calculated as:

`\begin{gathered} \sin\theta=\tan\theta\cdot\cos\theta \\ \sin\theta=(1)/(2)\cdot(-(2√(5))/(5)) \\ \sin\theta=-(√(5))/(5) \end{gathered}`
`\boxed{\sin\theta=-(√(5))/(5)}`