Need math help

[HybridSquirrel]

ok this is confusing for me

simplify

sin(pi/2 - x)

cot(5pi/2 - x)

csc(5pi + x)

 

[Schfifty Five]

$5 first.

 

[HybridSquirrel]

lol im poor

 

[Saint Nick]

Simplify? What do you mean simplify? Those are as simple as they get. At least I can't figure out anything else to do with them.

 

[HybridSquirrel]

i dunno the book just says simplify.

 

[Syringer]

Use these identities:

http://www.sosmath.com/trig/Trig5/trig5/trig5.html

 

[Saint Nick]

Use the first co-function identity given in the URL that syringer provided.

It looks like:

sin(pi/2 - theta) = cos theta.

I am guessing that the answer to the first one is along the lines of "cos x".

 

[HybridSquirrel]

kk ill try it with that link

 

[Syringer]

For example..

since sin (y-x) = sin x*cos y - cos x * sin y
sin(pi/2 - x)
= sin(pi/2)cos(x) - cos(pi/2) * sin x
sin pi/2 = 1
cos pi/2 = 0

So that is 1 * cos (x) - 0 * sin x
= cos(x)

 

[HybridSquirrel]

wow ok so use basically the same aproach for csc?

 

[QED]

You also need to remember that the sine and cosine functions are 2-Pi-periodic-- which means you can safely remove 2*Pi from any argument to a sine or cosine function without changing its value...

In other words:

sin(5*Pi / 2 - x) = sin( [5*Pi / 2 - x] - 2*Pi) = sin(Pi/2 - x), and similiarly
cos(5*Pi/ 2 - x) = cos(Pi/2 - x)

 

[QED]

Originally posted by: HybridSquirrel
wow ok so use basically the same aproach for csc?

Pretty much... but write out exactly what csc means, and simply THAT.

 

[manlymatt83]

I wish I had learned this stuff in high school when I was supposed to.

 

[ShadowBlade]

lemme grab my 89...

-sin(x-pi/2)
-1/(tan(x-(5pi/2)) ***Note: Domain of result may be larger
1/(sin(x+5pi))

TI-89 Titanium>*

 

[HybridSquirrel]

lol i have a ti-84

 

[matthekim]

Originally posted by: HybridSquirrel
lol i have a ti-84

lol

 

[HybridSquirrel]

i think its pretty nifty