# Help my life is in game

• Aug 18th 2010, 06:19 AM
TheGame
Help my life is in game
I need help with this (i know its easy but i dont know how to resolve this problem )
cosalfa = 5/13 sin alfa is - , cos alfa is + , tg alfa is - and ctg alfa is -

I need to find a) sin(60-alfa) b) cos(alfa +45)
i found out that sin alfa is 12/13

ps is degree

• Aug 18th 2010, 06:58 AM
mfetch22
Quote:

Originally Posted by TheGame
I need help with this (i know its easy but i dont know how to resolve this problem )
cosalfa = 5/13 sin alfa is - , cos alfa is + , tg alfa is - and ctg alfa is -

I need to find a) sin(60-alfa) b) cos(alfa +45)
i found out that sin alfa is 12/13

ps is degree

[Note that in the below I substituted x for your "alfa", so simplify things]

Your question is hard to understand, maybe try writing the problem in latex? Its not hard, its actually kinda cool when you get it to work. I kinda see your equations, but I dont understand the question because I am not certain what the equations are? Does "is" mean "=" ? I reccomend trying latex. Heres how you do it. You write these symbols (without the quotation marks, but including the "[" and "]" outside of the quotation marks) around any of your math stuff:

["math] equations go here [/math"]

Note that the last slash is a forward slash not a back slash, that messes me up all the time. Now, if you type in (without the quotation marks):

["math]cos(x) = 5/13 [/math"]

then you'll get:

$\displaystyle cos(x) = 5/13$

to make it allite more compact you can use a fraction, which is made by putting "\frac{top of fraction here}{bottom of fraction here}" into the "" symbols. So if you typed (without the quotations):

["math] cos(x) = \frac{5}{13} [/math"]

you'll get a nice pretty fraction like this:

$\displaystyle cos(x) = \frac{5}{13}$

Standard notation uses "tan(x)" to denote the tangent of x, not 'tg x'. It will just confuse people, just so you know for the furture to make things simpler on yourself. Same with 'ctg x'. Most people will not know that you meant "cot(x)". As for typing these into the math symbol magic maker, do the same thing, like this (without the quotations of course):

["math] sin(x) < 0 [/math"]

["math] cos(x) > 0 [/math"]

["math] tan(x) < 0 [/math"]

["math] cot(x) < 0 [/math"]

These give you:

$\displaystyle sin(x) < 0$

$\displaystyle cos(x) > 0$

$\displaystyle tan(x) < 0$

$\displaystyle cot(x) < 0$

I'm starting to understand what your question is. It took me awhile to realize what you were asking. Just so you know also, the best way to say "the tangent of x is negitive" is to say that "tan(x) < 0", as seen above. Now, if you don't mind, try rewriting your problem with these magic math symbols that make it look all fancy, and add any more details you may have left out, and also please provide the exact wording of the question in your textbook or worksheet or wherever, that is if you have the text book or worksheet or whatever the question was written on. Post that all back here, and I'll be happy to help you out with this problem. You might also want to take a quick look at this thread:

http://www.mathhelpforum.com/math-he...ns-153821.html

The problems in there sorta relate to what I think your problem is.
• Aug 18th 2010, 09:59 AM
Soroban
Hello, TheGame!

Quote:

$\displaystyle \cos\alpha = \frac{5}{13},\;\sin\alpha < 0$

Find:
. . $\displaystyle (a)\;\sin(60^o - \alpha)$
. . $\displaystyle (b)\;\cos(\alpha +45^o)$

Cosine is positive; $\displaystyle \alpha$ is in quadrant 1 or 4.
Sine is negative: $\displaystyle \alpha$ is in quadrant 3 or 4.

. . Hence: $\displaystyle \alpha$ is in quadrant 4.
. . And: .$\displaystyle \sin\alpha \:=\:\text{-}\frac{12}{13}$

$\displaystyle (a)\;\sin(60^o-\alpha) \;=\;\sin60^o\cos\alpha - \cos60^o\sin\alpha$

. . . . . . . . . . . . .$\displaystyle =\;\left(\dfrac{\sqrt{3}}{2}\right)\left(\dfrac{5} {13}\right) - \left(\dfrac{1}{2}\right)\left(\text{-}\dfrac{12}{13}\right)$

. . . . . . . . . . . . .$\displaystyle =\;\dfrac{5\sqrt{3} +12}{26}$

$\displaystyle (b)\;\cos(\alpha + 45^o) \;=\;\cos\alpha\cos45^o - \sin\alpha\sin45^o$

. . . . . . . . . . . . .$\displaystyle =\; \left(\dfrac{5}{13}\right)\left(\dfrac{1}{\sqrt{2} }\right) - \left(\text{-}\dfrac{12}{13}\right)\left(\dfrac{1}{\sqrt{2}}\ri ght)$

. . . . . . . . . . . . .$\displaystyle =\;\dfrac{5+12}{13\sqrt{2}} \;=\;\dfrac{17}{13\sqrt{2}}$