# Thread: cosine

1. ## cosine

It's one of those trigonometric ones that always make my brain go huh?

If cos x= (-4/5) and tan x>0, find the value of Sin x.

I figured that since Cos was negative, it was in either quadrants II or III. Since Tan is positive, it must be in III. After that....I get lost.

~Wynne

2. Originally Posted by Blodwynne
It's one of those trigonometric ones that always make my brain go huh?

If cos x= (-4/5) and tan x>0, find the value of Sin x.

I figured that since Cos was negative, it was in either quadrants II or III. Since Tan is positive, it must be in III. After that....I get lost.

~Wynne
Think of a right triangle. The cosine of an angle is the ratio of the adjacent side to the hypotenuse. So, ignoring the negative for now:

Code:
        /|
5   /  |
/    | b
/      |
/ θ      |
----------
4
By the Pythagorean theorem, we have

$\displaystyle b = \sqrt{25 - 16} = \sqrt9 = 3$ (this is a Pythagorean triple: all sides are integers).

Now, $\displaystyle \sin\theta = \frac{\text{opposite}}{\text{hypotenuse}}$, so we have

$\displaystyle \sin\theta = \frac b5 = \frac35$.

But, our $\displaystyle \theta$ is in quadrant I. Adding $\displaystyle 180^\circ$ to get into quadrant III will make the sine negative:

$\displaystyle \sin x = -\frac35$

3. Originally Posted by Blodwynne
It's one of those trigonometric ones that always make my brain go huh?

If cos x= (-4/5) and tan x>0, find the value of Sin x.

I figured that since Cos was negative, it was in either quadrants II or III. Since Tan is positive, it must be in III. After that....I get lost.

~Wynne
Draw a diagram

$\displaystyle \sin(x)=\frac{opposite}{hypotenuse}=\frac{-3}{5}$

Good luck

I'm too slow

4. Originally Posted by TheEmptySet
I'm too slow
Yes, but your graph with the triangle oriented the proper way is probably easier to understand than my explanation. Nice work.

5. Needless to say, that was far off from what I got. It turns out I did what I always do: I overthunk the problem. >.<

Sorry, but another trig problem is giving me issues. it is:

47. sin(cot^-1 5/12)

What I know so far:

Cotangent is the inverse of tangent.
This seems like another ratio of sides.

~Wynne

6. Originally Posted by Blodwynne
Needless to say, that was far off from what I got. It turns out I did what I always do: I overthunk the problem. >.<

Sorry, but another trig problem is giving me issues. it is:

47. sin(cot^-1 5/12)

What I know so far:

Cotangent is the inverse of tangent.
This seems like another ratio of sides.

~Wynne
Indeed it is. Set it up the same way:

$\displaystyle x = {\rm arccot}\frac5{12}\Rightarrow\cot x = \frac5{12}$

You know that $\displaystyle \cot\theta = \frac{\text{adjacent}}{\text{opposite}}$, so draw your triangle.

And since the cotangent is positive, and since the range of arccot is $\displaystyle [0,\;\pi]$ (or sometimes it is defined as $\displaystyle \left[-\frac\pi2,\;\frac\pi2\right]$), the angle you are dealing with ($\displaystyle x$) is in quadrant I.

Edit: I thought I should add: Be careful when you say that cotangent is the "inverse" of tangent. The word inverse usually implies an inverse function--e.g., arcsin is the inverse of sin. The proper word in your case is "reciprocal": the cotangent is the reciprocal of the tangent (reciprocal meaning a number's multiplicative inverse).

7. Originally Posted by Blodwynne
Needless to say, that was far off from what I got. It turns out I did what I always do: I overthunk the problem. >.<

Sorry, but another trig problem is giving me issues. it is:

47. sin(cot^-1 5/12)

What I know so far:

Cotangent is the inverse of tangent.
This seems like another ratio of sides.

~Wynne
Here is another diagram. I always draw them it helps.

We know that it has to be in quadrant I because the range of the arccot function is $\displaystyle [0,\pi]$ and the argument is positive

$\displaystyle \sin(\cos^{-1}\left( \frac{5}{12}\right))=\frac{12}{13}$

Edit: Too slow again. Twice in the same post