# Correct me if im wrong.. but

• December 21st 2006, 02:26 PM
3deltat
Correct me if im wrong.. but
The function of $cos \pi/6 =$ the cofunction of $sin \pi/6$ or something similar... (forgot how it was worded)

and would something like -2sin(4x)sin(-2x) be the same as 2sin(4x)sin(2x)
• December 21st 2006, 03:13 PM
ThePerfectHacker
Quote:

Originally Posted by 3deltat
The function of $cos \pi/6 =$ the cofunction of $sin \pi/6$ or something similar... (forgot how it was worded)

and would something like -2sin(4x)sin(-2x) be the same as 2sin(4x)sin(2x)

Sine and Cosine are Co-functions.
Meaning,
$\sin x=\cos \left( \frac{\pi}{2} -x \right)$
In the same way,
Tangent and Cotangent are cofunctions.
And,
Secant and Cosecant are cofunctions.

You can use this mnemonic to remember the properties of trigonometric cofunctions.
• December 21st 2006, 03:31 PM
3deltat
Quote:

Sine and Cosine are Co-functions.
Meaning,

In the same way,
Tangent and Cotangent are cofunctions.
And,
Secant and Cosecant are cofunctions.
Ah. So the statement "The function of $cos \pi/6$ Would = the cofunction of $sin \pi/6$ ? =)
• December 21st 2006, 05:06 PM
ThePerfectHacker
Quote:

Originally Posted by 3deltat
Ah. So the statement "The function of $cos \pi/6$ Would = the cofunction of $sin \pi/6$ ? =)

Thus, the by the cofunction identities,
$\cos \frac{\pi}{6}=\sin \left( \frac{\pi}{2} - \frac{\pi}{6} \right)=\sin \frac{\pi}{3}$
• December 22nd 2006, 05:47 AM
topsquark
Quote:

Originally Posted by 3deltat
would something like -2sin(4x)sin(-2x) be the same as 2sin(4x)sin(2x)

As sine is an odd function $sin(-x) = -sin(x)$. Thus
$-2sin(4x)sin(-2x) = -2sin(4x) \cdot -sin(2x) = 2sin(4x)sin(2x)$.