# Trig help!

• Aug 11th 2006, 09:24 PM
rain_lover
Trig help!
Hey guys im abit confused with the different quadrants of the trig ratios thing.

I know all trig ratios are positive in 1st quadrant, then sine, tan, cosine only. However, why does:

cos x= -5/11 (solve for x if 0<x<180)
= 117 degrees?
I thought only sine could be positive when greater than 90 degrees and less than 180 (obtuse). Does - sign mean that this negates this rule?

Also, what would happen if sin x= -5/11 (solve for x if 0<x<180)?

I get two answers for x:
x= 153 degrees or 27 degrees. Since sine is positive both in acute and obtuse quadrants, does this mean both answers are valid?

Any help is much appreciated
• Aug 11th 2006, 10:14 PM
CaptainBlack
Quote:

Originally Posted by rain_lover
Hey guys im abit confused with the different quadrants of the trig ratios thing.

I know all trig ratios are positive in 1st quadrant, then sine, tan, cosine only. However, why does:

cos x= -5/11 (solve for x if 0<x<180)
= 117 degrees?
I thought only sine could be positive when greater than 90 degrees and less than 180 (obtuse). Does - sign mean that this negates this rule?

This illustrates the point nicely. you are told that $\displaystyle \cos(x)$ is negative, which means
that $\displaystyle x$ is in quadrant 2 and/or 3, since $\displaystyle \cos(x)$ is positive only in quadrants 1 and 4.

The angle in quadrant 2 for which $\displaystyle \cos(x)=-5/11$ is $\displaystyle \approx 117^{\circ}$, and in quadrant 3
is $\displaystyle \approx 243^{\circ}$

RonL
• Aug 11th 2006, 10:43 PM
rain_lover
Hey thanks but how did you get 243 degrees for the one in the 3rd quadrant?? Also, how would you find sin x = -5/11? Would you get 1 or 2 answers?

Thanks
• Aug 12th 2006, 02:34 AM
CaptainBlack
Quote:

Originally Posted by rain_lover
Hey thanks but how did you get 243 degrees for the one in the 3rd quadrant?? Also, how would you find sin x = -5/11? Would you get 1 or 2 answers?

Thanks

The 243 degrees comes from knowing that the other angle is 360-(180-117).
That is by knowing the symmetry properties of the trig functions.

It should be in your notes/text.

RonL
• Aug 12th 2006, 02:38 AM
Quick
Quote:

Originally Posted by rain_lover
Also, how would you find sin x = -5/11? Would you get 1 or 2 answers?

Thanks

Look for yourself: (the graph is in degrees)
• Aug 19th 2006, 11:27 AM
Soroban
Hello, rain_lover!

Your questions are confusing to me . . .

Quote:

I know all trig ratios are positive in 1st quadrant, then sine, tan, cosine only.
However, why does: .$\displaystyle \cos x= -\frac{5}{11} = 117^o$ ?

I thought only sine could be positive when 90° < x < 180° (obtuse).

This is true . . . but why do you care? . . . This problem involves the cosine.

Quote:

Does - sign mean that this negates this rule?

No, the rule for sine still holds . . . why wouldn't it?

Quote:

Also, what would happen if $\displaystyle \sin x= -\frac{5}{11}$ (solve for 0 < x < 180) ?

Quite impossible . . .

As you pointed out, sine is positive in Quadrant 1 and 2 (0 < x < 180).
So, x cannot be negative there.

• Aug 19th 2006, 02:50 PM
spiritualfields
Quote:

The 243 degrees comes from knowing that the other angle is 360-(180-117).
That is by knowing the symmetry properties of the trig functions.
Since the x/h value is -5/11, indicating that the angles are in the second and third quadrants, shouldn't the other angle (besides 117) be figured by adding 63 degrees to 180 degrees? Or by subtracting 117 from 360?

Just wanted to clear this up for the original poster, who seems to be having a hard time understanding this.
• Aug 19th 2006, 09:33 PM
CaptainBlack
Quote:

Originally Posted by spiritualfields
Since the x/h value is -5/11, indicating that the angles are in the second and third quadrants, shouldn't the other angle (besides 117) be figured by adding 63 degrees to 180 degrees? Or by subtracting 117 from 360?

Just wanted to clear this up for the original poster, who seems to be having a hard time understanding this.

Since they are the same thing you do whatever seems more natural to
you.

RonL
• Aug 20th 2006, 09:35 AM
spiritualfields
But 360 - (180 - 117) = 297, which is the fourth quadrant.
• Aug 20th 2006, 09:47 AM
CaptainBlack
Quote:

Originally Posted by spiritualfields
But 360 - (180 - 117) = 297, which is the fourth quadrant.

What does this refer to? The last post in this thread was:

Quote:

Originally Posted by CaptainBlack
Quote:

Originally Posted by spiritualfields
Since the x/h value is -5/11, indicating that the angles are in the second and third quadrants, shouldn't the other angle (besides 117) be figured by adding 63 degrees to 180 degrees? Or by subtracting 117 from 360?

Just wanted to clear this up for the original poster, who seems to be having a hard time understanding this.

Since they are the same thing you do whatever seems more natural.

The comment refers to 180+63 degrees being the same as 360-117.

RonL
• Aug 20th 2006, 10:20 AM
spiritualfields
My comments are with respect to posts # 1, 2, and 3, refering specifically to the cos x = -5/11, and the op's question about how could x be 117 degrees as well as 243 degrees. The way I arrive at the 243 is by adding 63 to 180 (because I know that 180 - 117 = 63) or by subtracting 117 from 360. You get to the same angle (243) either way. I'm no math whiz, but THIS much I know! Given that, and the context of posts 1, 2, and 3, I'm not sure what 360 - (180 - 117) is supposed represent. If I were a student in your class, I'd be calling you out on that one ;)
• Aug 20th 2006, 10:27 AM
CaptainBlack
Quote:

Originally Posted by spiritualfields
My comments are with respect to posts # 1, 2, and 3, refering specifically to the cos x = -5/11, and the op's question about how could x be 117 degrees as well as 243 degrees. The way I arrive at the 243 is by adding 63 to 180 (because I know that 180 - 117 = 63) or by subtracting 117 from 360. You get to the same angle (243) either way. I'm no math whiz, but THIS much I know! Given that, and the context of posts 1, 2, and 3, I'm not sure what 360 - (180 - 117) is supposed represent. If I were a student in your class, I'd be calling you out on that one ;)

You should try to quote at least some context, otherwise the thread
can become confusing and difficult to follow.

There are tricks that will allow you to quote from multiple previous posts
if you need to (Use two browser windows, with the post you are creating
in one window. Open the same thread in the other window and click the
quote button on any post you want to quote, select the text in the reply
box then <control>c will copy the text onto the clipboard. Change back to
the first window then <control>v will paste the text from the clipboard into
the current reply box. Repeat for any other posts you need to quote).

RonL