# Equivalent Trigonomteric Expressions

• Mar 29th 2009, 08:38 AM
skeske1234
Equivalent Trigonomteric Expressions
Can someone explain to me exactly why cosx=sin(pi/2-x) and why sinx=cos(pi/2-x)

Also why are tanx=cot(pi/2-x) and cotx=tan(pi/2-x)?

I don't understand why they are related? how does cosx=sin(pi/2-x) exactly?
My teacher just kind of gave us the formulas.. but never said WHY they're like that, she didn't prove it or anything.. Can someone please prove it or explain the concept to me.

Thank you
• Mar 29th 2009, 08:43 AM
Math Major
Well, start by writing out a table for 0-pi/2 for sine and cosine.

Sin(x)

sin(0) = 0
sin(pi/6) = 1/2
sin(pi/4) = 1/root(2)
sin(pi/3) = root(3)/2
sin(pi/2) = 1

Cos(x)
cos(0) = 1
cos(pi/6) = root(3)/2
cos(pi/4) = 1/root(2)
cos(pi/3) = 1/2
cos(pi/2) = 0

Notice anything in particular about them?
• Mar 29th 2009, 08:51 AM
skeeter
note that cosine is the sine of an angle's complement. in other words, cosine is short for "complementary sine"

same idea for the other "co-functions", cotangent and cosecant.
• Mar 29th 2009, 09:04 AM
skeske1234
how would you sketch the diagram (triangle) to show this or prove this for the cotangent and tangent angles and the secant and cosecant?

I don't get those ones still
• Mar 29th 2009, 09:22 AM
skeeter
Quote:

Originally Posted by skeske1234
how would you sketch the diagram (triangle) to show this or prove this for the cotangent and tangent angles and the secant and cosecant?

I don't get those ones still

let "A" and "B" be the two acute angles in a right triangle, with sides opposite them "a" and "b" respectively ...

$\tan{A} = \frac{a}{b} = \cot{B}$

$\cot{A} = \frac{b}{a} = \tan{B}$

same argument with secant and cosecant