# Proving Indentity

• Mar 16th 2006, 01:05 PM
milkywayio
Proving Indentity

1. cos 2 x = cotx - tanx
sinx cosx

2. (cosx-sinx)^2 = 1 - sin 2 x

3. secx - sinxtanx = cosx

4. cos(x+y)cosy + sin(x+y) sin y = cosx
• Mar 16th 2006, 01:20 PM
ThePerfectHacker
Quote:

Originally Posted by milkywayio

1. cos 2 x = cotx - tanx
sinx cosx

For Left Hand Side
$\displaystyle \frac{\cos 2x}{\sin x\cos x}$
Express as,
$\displaystyle \frac{2\cos 2x}{2\sin x\cos x}$
Double angle for,
$\displaystyle \frac{2\cos 2x}{\sin 2x}$
Thus,
$\displaystyle 2\cot 2x$

For Right Hand Side,
$\displaystyle \cot x-\tan x$
Use reciprocal identities to get,
$\displaystyle \frac{\cos x}{\sin x}-\frac{\sin x}{\cos x}$
$\displaystyle \frac{\cos^2x-\sin^2x}{\sin x\cos x}$
Use double angle for cosine,
$\displaystyle \frac{\cos 2x}{\sin x\cos x}$
Express as,
$\displaystyle \frac{2\cos 2x}{2\sin x\cos x}$
Use double angle for sine,
$\displaystyle \frac{2\cos 2x}{\sin 2x}$
Use identities,
$\displaystyle 2\cot 2x$
Thus, the left hand and the right hand give the same result
Q.E.D.
• Mar 16th 2006, 01:23 PM
ThePerfectHacker
Quote:

Originally Posted by milkywayio

2. (cosx-sinx)^2 = 1 - sin 2 x

Open the Left hand side remember to use FOIL,
$\displaystyle (\cos x-\sin x)^2=\cos^2x-2\sin x\cos x+\sin^2x$
Notice, that
$\displaystyle \cos^2x+\sin^2x=1$ and $\displaystyle -2\sin x \cos x=-2\sin x$
Thus,
$\displaystyle (\cos x-\sin x)^2=1-\sin 2x$
Q.E.D.
• Mar 16th 2006, 01:28 PM
ThePerfectHacker
Quote:

Originally Posted by milkywayio
3. secx - sinxtanx = cosx

Express tangent through sine and cosine,
$\displaystyle \sec x-\sin x\cdot \frac{\sin x}{\cos x}$
Express, secant through cosine,
$\displaystyle \frac{1}{\cos x}-\sin x\cdot \frac{\sin x}{\cos x}$
$\displaystyle \frac{1-\sin^2x}{\cos x}$
Notice that $\displaystyle 1-\sin^2x=\cos^2x$
Thus,
$\displaystyle \frac{\cos^2x}{\cos x}=\cos x$
Thus,
$\displaystyle \sec x-\tan x\sin x=\cos x$
Q.E.D.
• Mar 16th 2006, 01:32 PM
ThePerfectHacker
Quote:

Originally Posted by milkywayio
4. cos(x+y)cosy + sin(x+y) sin y = cosx

Call, $\displaystyle z=x+y$
Thus,
$\displaystyle \cos z\cos y+\sin z\sin y$
Notice that this is the cosine for difference of two angles, thus,
$\displaystyle \cos (z-y)$
Substitute that value of $\displaystyle z$ thus,
$\displaystyle \cos (x+y-y)=\cos x$
Thus,
$\displaystyle \cos (x+y)\cos y + \sin (x+y) \sin y=\cos x$
Q.E.D.