Trig identities, exact values

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November 4th 2007, 07:26 PM
zerostumbleine33

Trig identities, exact values

Hey, I was absent the day my class went over how to solve these problems. So I was wondering if I could get some help.

-Use the fundamental identities to find the exact values of the remaining trig functions of x...

1)cos x= square root 7/4 and cot x= - square root 7/3

Would the sec x= 1/square root 7/4...? Or is that not the way to think of it?

-Simplify each expression using the fundamental identities

2) Sec theta X cos theta

3) 1/csc^2x + 1/sec^2x

4) 1-sin^2u/cosu

5) cos^2x+ (sin x +1)^2/sin x + 1

Thank you so much if you can help me. Also, is there a chat room or anything of the sort on this site, I'm new.
:)

-and yes, I'm horrible at math haha
November 4th 2007, 07:38 PM
topsquark

Quote:

Originally Posted by zerostumbleine33

1)cos x= square root 7/4 and cot x= - square root 7/3

I presume these are
$cos(x) = \frac{\sqrt{7}}{4}$
and
$cot(x) = - \frac{\sqrt{7}}{3}$

We know that cosine is positive and cotangent is negative. Where does this happen? In Quadrant IV. Thus we know where the angle is.

We want $sin(x)$ . Well, we know that $sin^2(x) + cos^2(x) = 1$ , so we can get that easily enough. (Note that sine is negative in QIV.)

Getting $tan(x)$ is pretty easy:
$tan(x) = \frac{1}{cot(x)} = -\frac{3}{\sqrt{7}}$

Now the last two are also easy:
$csc(x) = \frac{1}{sin(x)}$
and
$sec(x) = \frac{1}{cos(x)}$

-Dan

Edit: I just thought of another way to get $sin(x)$ :
$sin(x) = \frac{cos(x)}{cot(x)}$
This is a faster method, if you know how to deal with the complex fraction.
November 5th 2007, 03:04 PM
zerostumbleine33

In the first equation, the square root goes over the 7/4, same with the - being outside the square root over 7/3...so that would still make it in QIV, but what else would that affect?
November 6th 2007, 03:50 AM
topsquark

Quote:

Originally Posted by zerostumbleine33

In the first equation, the square root goes over the 7/4, same with the - being outside the square root over 7/3...so that would still make it in QIV, but what else would that affect?

Nothing really. You will have to do the problem over again using the correct values, but the method is still the same.

-Dan