Finding cos θ when sin θ = -5/13 and θ is in the fourth quadrant

[CrossFit415]

Sin $$\theta$$ = -5/13,
(3$$\pi$$ / 2) < $$\theta$$ < 2 $$\pi$$

So I got cos $$\theta$$ = - (sqrt 194) / 13
Is this the right answer?

 

[icystrike]

How u manage to get that? Its incorrect

 

[CrossFit415]

I used sin^2theta + cos^2theta = 1

 

[eumyang]

And I bet that when you did, you squared -5/13 and got -25/169? If so, that's not right -- it's +25/169.

Also, what quadrant is
$$\frac{3\pi}{2} < \theta < 2\pi$$
? And what is the sign for the cosine ratio in that quadrant?

Finally, please don't use LaTex for single characters. It doesn't look right. Click the inequality I wrote above to see how to type it.

 

[CrossFit415]

Cos theta = -12/13 . And (-) since it's in quadrant IV.

 

[CrossFit415]

Thanks

 

[eumyang]

Cos theta = -12/13 . And (-) since it's in quadrant IV.

No, no, no, cosine is positive in Q IV. You really need to remember the signs of the trig ratios in each quadrant.

Q I: All
Q II: sine (and cosecant)
Q III: tangent (and cotangent)
Q IV: cosine (and secant)

When I learned this, I was taught the mnemonic "All Students Take Calculus" to help me remember.