1. ## 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

2. Originally Posted by milkywayio

1. cos 2 x = cotx - tanx
sinx cosx
For Left Hand Side
$\frac{\cos 2x}{\sin x\cos x}$
Express as,
$\frac{2\cos 2x}{2\sin x\cos x}$
Double angle for,
$\frac{2\cos 2x}{\sin 2x}$
Thus,
$2\cot 2x$

For Right Hand Side,
$\cot x-\tan x$
Use reciprocal identities to get,
$\frac{\cos x}{\sin x}-\frac{\sin x}{\cos x}$
$\frac{\cos^2x-\sin^2x}{\sin x\cos x}$
Use double angle for cosine,
$\frac{\cos 2x}{\sin x\cos x}$
Express as,
$\frac{2\cos 2x}{2\sin x\cos x}$
Use double angle for sine,
$\frac{2\cos 2x}{\sin 2x}$
Use identities,
$2\cot 2x$
Thus, the left hand and the right hand give the same result
Q.E.D.

3. Originally Posted by milkywayio

2. (cosx-sinx)^2 = 1 - sin 2 x
Open the Left hand side remember to use FOIL,
$(\cos x-\sin x)^2=\cos^2x-2\sin x\cos x+\sin^2x$
Notice, that
$\cos^2x+\sin^2x=1$ and $-2\sin x \cos x=-2\sin x$
Thus,
$(\cos x-\sin x)^2=1-\sin 2x$
Q.E.D.

4. Originally Posted by milkywayio
3. secx - sinxtanx = cosx
Express tangent through sine and cosine,
$\sec x-\sin x\cdot \frac{\sin x}{\cos x}$
Express, secant through cosine,
$\frac{1}{\cos x}-\sin x\cdot \frac{\sin x}{\cos x}$
$\frac{1-\sin^2x}{\cos x}$
Notice that $1-\sin^2x=\cos^2x$
Thus,
$\frac{\cos^2x}{\cos x}=\cos x$
Thus,
$\sec x-\tan x\sin x=\cos x$
Q.E.D.

5. Originally Posted by milkywayio
4. cos(x+y)cosy + sin(x+y) sin y = cosx
Call, $z=x+y$
Thus,
$\cos z\cos y+\sin z\sin y$
Notice that this is the cosine for difference of two angles, thus,
$\cos (z-y)$
Substitute that value of $z$ thus,
$\cos (x+y-y)=\cos x$
Thus,
$\cos (x+y)\cos y + \sin (x+y) \sin y=\cos x$
Q.E.D.