# Simplifying Trigonometric Equations

• Feb 23rd 2009, 01:52 PM
sar
Simplifying Trigonometric Equations
Hi all! I know this is probably a fairly simple concept for Trigonometry, but I can't quite remember how to do them. I'm just trying to make sure I know them well enough so that I can do them on a test I have in a few days. It would be awesome if you could just get me started on them.

1. $\displaystyle cot(x)sin(x) = \frac{\sqrt{3}}{2}$

2. $\displaystyle tan(x)$ • $\displaystyle cos(x) = 1$

3. $\displaystyle sin^2(x)$ + $\displaystyle cos^2(x) + cos(x)$ = 1

4. [$\displaystyle cot(x)$ • $\displaystyle (\frac{sin(x)}{cos(x)})$] • $\displaystyle sec^2(x)$ = $\displaystyle \frac{1}{2}$
• Feb 23rd 2009, 02:46 PM
wytiaz
Quote:

Originally Posted by sar
Hi all! I know this is probably a fairly simple concept for Trigonometry, but I can't quite remember how to do them. I'm just trying to make sure I know them well enough so that I can do them on a test I have in a few days. It would be awesome if you could just get me started on them.

1. $\displaystyle cot(x)sin(x) = \frac{\sqrt{3}}{2}$

It is normally best to write all of your sec(x), csc(x), tan(x), and cot(x) in terms of sin(x) and cos(x). That way, if anything cancels, you can simplify the expression.

$\displaystyle \frac{cos(x)}{sin(x)} • sin(x) = \frac{\sqrt{3}}{2}$

Hey, there's a sin(x) on the top as well as the bottom! So, as long as the sin(x) is not zero, we're ok to divide the top and bottom by sin(x) and cancel them.

$\displaystyle cos(x) = \frac{\sqrt{3}}{2}$

So which angles have a cosine value of $\displaystyle \frac{\sqrt{3}}{2}$ ?

Once you've deduced what x is, don't forget, you can't have any angles that would let sin(x) be equal to zero. So you'll have to examine your answers and refuse any answers that had a sin value of zero.
• Feb 23rd 2009, 03:32 PM
sar
Wow, that made it so much easier! Thanks a bunch! (Rock)