Question

Angle LaTeX: \theta θ has a terminal side located in Quadrant II of the coordinate plane and LaTeX: \sin\theta=\frac{5}{13} sin θ = 5 13 . What is LaTeX: \cos\theta cos θ ?

Answer 1

`\(\cos \theta = -(12)/(13)\)` when
`\(\sin \theta = (5)/(13)\)` and
`\(\theta\)`is in Quadrant II.

In Quadrant II, the sine function is positive. You are given that
`\(\sin \theta = (5)/(13)\)`.

Using the Pythagorean identity for sine and cosine, which states that
`\(\sin^2 \theta + \cos^2 \theta = 1\)`, you can find
`\(\cos \theta\)`:

`\[\cos^2 \theta = 1 - \sin^2 \theta\]`

Plug in the given value of
`\(\sin \theta\)`:

`\[\cos^2 \theta = 1 - \left((5)/(13)\right)^2\]`

Now, solve for
`\(\cos \theta\)`:

`\[\cos \theta = \pm \sqrt{1 - \left((5)/(13)\right)^2}\]`

`\[\cos \theta = \pm \sqrt{1 - (25)/(169)}\]`

`\[\cos \theta = \pm \sqrt{(144)/(169)}\]`

`\[\cos \theta = \pm (12)/(13)\]`

Since
`\(\theta\)` is in Quadrant II, where the cosine function is negative, choose the negative sign:

`\[\cos \theta = -(12)/(13)\]`

So,
`\(\cos \theta = -(12)/(13)\)` when
`\(\sin \theta = (5)/(13)\)` and
`\(\theta\)`is in Quadrant II.

The probable question may be:

"In the second quadrant of the coordinate plane, the cosine function is negative. Therefore, if
`\(\sin\theta = (5)/(13)\)`, the cosine function is given by
`\(\cos\theta = -√(1 - \sin^2\theta)\).`

Substitute the given value of
`\(\sin\theta\)`:

`\[\cos\theta = -\sqrt{1 - \left((5)/(13)\right)^2}\]`

Calculate the expression inside the square root, and then take the negative square root of that result to find
`\(\cos\theta\)`."