# cosine

• Jun 8th 2008, 09:04 PM
Blodwynne
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
• Jun 8th 2008, 09:15 PM
Reckoner
Quote:

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:

$\displaystyle \sin x = -\frac35$
• Jun 8th 2008, 09:18 PM
TheEmptySet
Quote:

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

Attachment 6703

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

Good luck

I'm too slow
• Jun 8th 2008, 09:22 PM
Reckoner
Quote:

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.
• Jun 8th 2008, 09:25 PM
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
• Jun 8th 2008, 09:47 PM
Reckoner
Quote:

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).
• Jun 8th 2008, 09:50 PM
TheEmptySet
Quote:

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

Attachment 6705

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

Edit: Too slow again. Twice in the same post :D