Question

Find cos θ if sin θ = -12/13 and tan θ > 0.

MEDAL REWARDED!

Answer 1

Use the fundamental identity, \(\sin^2\theta + \cos^2 \theta=1\) to solve for \(\cos\theta\).

That will get you to with a \(\pm\).

Then use the fact that sine is negative and tangent is positive to determine what quadrant \(\theta\) is in - that will give you the correct sign for (\cos\theta\).

Answer 2

Can you just give me the answer and then how you found it? I find it much easier to learn that way and this is just a practice problem so its not worth anything.

Answer 3

\(\bf sin(\theta) = -\cfrac{12}{13}\implies\cfrac{\textit{opposite}}{\textit{hypotenuse}}\implies \cfrac{b}{c}\\ \quad \\
\textit{what's }\quad a?\\
c^2 = a^2 + b^2\implies \pm\sqrt{c^2-b^2}=a\)

keep in mind that, the tangent of the angle, is >0, meaning is positive
\(\bf tan(\theta) = \cfrac{sin(\theta)}{cos(\theta)}\) so, the only way for the tangent to be positive, is if both, sine and cosine are positive, or both sine and cosine are negative

notice above, your sine given is negative, thus the cosine, and therefore "a", is negative too

Answer 4

keep in mind that \(\bf cos(\theta) = \cfrac{\textit{adjacent}}{\textit{hypotenuse}}\implies \cfrac{a}{c}\)

Answer 5

Which one would i be?
-5/12
-5/13
12/5
-13/12

Answer 6

well... what did you get for "a"?

Answer 7

I have trouble putting everything into the formulas, thats why i was asking if someone could do this problem for me and show me the answer and how they got it, i have other problems im going to do on my own

Answer 8

\(\bf sin(\theta) = -\cfrac{12}{13}\implies\cfrac{\textit{opposite}}{\textit{hypotenuse}}\implies \cfrac{b}{c}\\ \quad \\
\textit{what's }\quad a?\\
c^2 = a^2 + b^2\implies \pm\sqrt{c^2-b^2}=a\\
sin(\theta) = -\cfrac{12}{13}\implies \cfrac{b}{c}\implies b = -12\qquad c = 13\\ \quad \\
\pm\sqrt{13^2-(-12)^2}=a\implies a = \pm\sqrt{169-144}\\ \quad \\ a = \pm\sqrt{25}\implies a = -5
\)

Answer 9

@jdoe0001 okay, and then from there?

Answer 10

well.. \(\bf cos(\theta) = \cfrac{\textit{adjacent}}{\textit{hypotenuse}}\implies \cfrac{a}{c}\)

Answer 11

-5/13!