Proving Indentity

milkywayio · March 16th 2006, 01:05 PM

i have no idea how to do these four problems please help me!!!

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

**ThePerfectHacker** · March 16th 2006, 01:20 PM

Originally Posted by milkywayio

i have no idea how to do these four problems please help me!!!

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,
$\frac{\cos x}{\sin x}-\frac{\sin x}{\cos x}$
Add fractions,
$\frac{\cos^2x-\sin^2x}{\sin x\cos x}$
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.

**ThePerfectHacker** · March 16th 2006, 01:23 PM

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.

**ThePerfectHacker** · March 16th 2006, 01:28 PM

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,
$\frac{1}{\cos x}-\sin x\cdot \frac{\sin x}{\cos x}$
Thus, add fractions
$\frac{1-\sin^2x}{\cos x}$
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.

**ThePerfectHacker** · March 16th 2006, 01:32 PM

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.

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