How do you simplify this problem?

Darkhrse99 · #1 $Old$ June 18th, 2008, 03:01 PM

$sin(\frac{-3pi}{4}) cos(\frac{5pi}{6})$

Thanks

TKHunny · #2 $Old$ June 18th, 2008, 03:12 PM

Quote:

Originally Posted by Darkhrse99 $View Post$

$sin(\frac{-3pi}{4}) cos(\frac{5pi}{6})$

Didn't I tell you before to memorize certain trigonometric values? It seems you have not fulfilled this assignment.

sin(-3pi/4) = -sin(3pi/4) = -sin(pi/4) = ???

cos(5pi/6) = -cos(pi/6) = ???

Now what?

Note: If you're using LaTeX, feel free to put a slash in front of those "pi"s. Excellent results can be obtained.

Darkhrse99 · #3 $Old$ June 18th, 2008, 03:18 PM

Quote:

Originally Posted by TKHunny $View Post$

Didn't I tell you before to memorize certain trigonometric values? It seems you have not fulfilled this assignment.

sin(-3pi/4) = -sin(3pi/4) = -sin(pi/4) = ??? $\frac{\sqrt1}{2}$ ?

cos(5pi/6) = -cos(pi/6) = ??? $\frac{1}{2}$ ?

Now what?

Note: If you're using LaTeX, feel free to put a slash in front of those "pi"s. Excellent results can be obtained.

Is this correct?

TKHunny · #4 $Old$ June 18th, 2008, 04:28 PM

Sorry, it's a little like you are just throwing darts.

On the first:

1) Where did the negative sign go?

2) $\sqrt{1}$ ?

On the second:

3) Where did the negative sign go?

4) Please note the difference between $\frac{\pi}{6}$ and $\frac{\pi}{3}$ .

Look at a chart of values. Think it through. Don't just stab at it.

Darkhrse99 · #5 $Old$ June 18th, 2008, 05:24 PM

So the first on is $\frac{-\sqrt1}{2}$ and the second one is. $\frac{-\sqrt3} {2}$ . I'm finding the (-)(Y value for $\frac{\pi}{4}$
and the X value for $\frac{\pi}{4}$ Is this correct. I'm not trying to take stabs at it, but i truely don't know how to do the problem.

Darkhrse99 · #6 $Old$ June 18th, 2008, 07:51 PM

Is this answer correct?

TKHunny · #7 $Old$ June 18th, 2008, 08:00 PM

You must MEMORIZE a few important values.

$0\;\;\sin(0)\;=\;0\;\;\cos(0)\;=\;1$

$\frac{\pi}{6}\;\;\sin(\frac{\pi}{6})\;=\;\frac{1}{2}\;\;\cos(\frac{\pi}{6})\;=\;\frac{\sqrt{3}}{2}$

$\frac{\pi}{4}\;\;\sin(\frac{\pi}{4})\;=\;\frac{\sqrt{2}}{2}\;\;\cos(\frac{\pi}{4})\;=\;\frac{\sqrt{2}}{2}$

$\frac{\pi}{3}\;\;\sin(\frac{\pi}{3})\;=\;\frac{\sqrt{3}}{2}\;\;\cos(\frac{\pi}{3})\;=\;\frac{1}{2}$

$\frac{\pi}{2}\;\;\sin(\frac{\pi}{2})\;=\;1\;\;\cos(\frac{\pi}{2})\;=\;0$

There is NO substitute for memorizing this tableau. Put it in your brain.

Note: You still have $\sqrt{1}$ . Don't write things that make no sense.

Darkhrse99 · #8 $Old$ June 18th, 2008, 08:12 PM

Thanks for the help.

My text book shows that $\frac{\sqrt1}{2}$ is the same as $\frac{\sqrt2}{2}$

Darkhrse99 · #9 $Old$ June 18th, 2008, 08:29 PM

So really what I'm trying to do when simplifying is getting the x or y of the above table?

So in the problem below I need to find the X or Y value?
Simplify $tan(\frac{-2\pi}{3})csc(\frac{3\pi}{4})$
Tan= $\frac{Y}{X}$ CSC= $\frac{1}{Y}$ Do I find the x or y value for tan and cosecant?

Moo · #10 $Old$ June 19th, 2008, 02:54 AM

Quote:

Originally Posted by Darkhrse99 $View Post$

Thanks for the help.

My text book shows that $\frac{\sqrt1}{2}$ is the same as $\frac{\sqrt2}{2}$

No, once again, it's not $\frac{\sqrt1}{2}$ but $\sqrt{\frac 12}$ which is the same as $\frac{\sqrt2}{2}$

TKHunny · #11 $Old$ June 19th, 2008, 06:51 AM

or maybe $\frac{1}{\sqrt{2}}$ .

Notation means stuff. Don't disregard it for expediency.

TKHunny · #12 $Old$ June 19th, 2008, 06:53 AM

Quote:

Originally Posted by Darkhrse99 $View Post$

So really what I'm trying to do when simplifying is getting the x or y of the above table?

So in the problem below I need to find the X or Y value?
Simplify $tan(\frac{-2\pi}{3})csc(\frac{3\pi}{4})$
Tan= $\frac{Y}{X}$ CSC= $\frac{1}{Y}$ Do I find the x or y value for tan and cosecant?

What does your memorized table tell you?