Differentiation's Proof ..

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February 18th 2010, 09:25 AM
Ted

Differentiation's Proof ..

Show that:
$\frac{d}{dx} \left( \frac{sin^2(x)}{1+cot(x)} + \frac{cos^2(x)}{1+tan(x)} \right) = -cos(2x)$ .

Am sure, there is a trick somewhere.
sure, we will not differentiate it normally.
I rewrite the function as:
February 18th 2010, 09:33 AM
Calculus26

Quote:

http://www.mathhelpforum.com/math-he...956fc68b-1.gif

the function reduces to 1- 1/2sin(2x)

convert to sines and cosines and combine into a single term

recall a^3 + b^3 = (a+b) (a^2 -ab +b^2)
February 18th 2010, 09:55 AM
Soroban

Hello, Ted!

You're right . . . We're expected to simplify the function first.

Quote:

$\text{Show that: }\;\frac{d}{dx}\left( \frac{\sin^2x}{1+\cot x} + \frac{\cos^2x}{1+\tan x } \right) \;=\; -\cos 2x$ .

We have: . $\frac{\sin^2\!x}{1 + \frac{\cos x}{\sin x}} + \frac{\cos^2\!x}{1 + \frac{\sin x}{\cos x}}$

. . . . . . $=\;{\color{blue}\frac{\sin x}{\sin x}}\cdot\frac{\sin^2\!x}{1 + \frac{\cos x}{\sin x}} \;+\; {\color{blue}\frac{\cos x}{\cos x}}\cdot\frac{\cos^2\!x}{1 + \frac{\sin x}{\cos x}}$

. . . . . . $=\;\frac{\sin^3\!x}{\sin x + \cos x} + \frac{\cos^3\!x}{\sin x + \cos x}$

. . . . . . $=\;\frac{\sin^3\!x + \cos^3\!x}{\sin x + \cos x}$

. . . . . . $=\;\frac{(\sin x + \cos x)(\sin^2\!x - \sin x\cos x + \cos^2\!x)}{\sin x +\cos x}$

. . . . . . $=\;\underbrace{\sin^2\!x + \cos^2\!x}_{\text{This is 1}} - \sin x\cos x$

. . . . . . $=\; 1 - \sin x\cos x$

. . . . . . $=\;1 - \tfrac{1}{2}\sin2x$

Now differentiate it!

Edit: Ah, Calculus 26 beat me to it . . . *sigh*
.
February 18th 2010, 10:10 AM
vince

Quote:

Originally Posted by Soroban

Hello, Ted!

You're right . . . We're expected to simplify the function first.

We have: . $\frac{\sin^2\!x}{1 + \frac{\cos x}{\sin x}} + \frac{\cos^2\!x}{1 + \frac{\sin x}{\cos x}}$

. . . . . . $=\;{\color{blue}\frac{\sin x}{\sin x}}\cdot\frac{\sin^2\!x}{1 + \frac{\cos x}{\sin x}} \;+\; {\color{blue}\frac{\cos x}{\cos x}}\cdot\frac{\cos^2\!x}{1 + \frac{\sin x}{\cos x}}$

. . . . . . $=\;\frac{\sin^3\!x}{\sin x + \cos x} + \frac{\cos^3\!x}{\sin x + \cos x}$

. . . . . . $=\;\frac{\sin^3\!x + \cos^3\!x}{\sin x + \cos x}$

. . . . . . $=\;\frac{(\sin x + \cos x)(\sin^2\!x - \sin x\cos x + \cos^2\!x)}{\sin x +\cos x}$

. . . . . . $=\;\underbrace{\sin^2\!x + \cos^2\!x}_{\text{This is 1}} - \sin x\cos x$

. . . . . . $=\; 1 - \sin x\cos x$

. . . . . . $=\;1 - \tfrac{1}{2}\sin2x$

Now differentiate it!

Edit: Ah, Calculus 26 beat me to it . . . *sigh*
.

the colors the indenting... $\;\underbrace{\sin^2\!x + \cos^2\!x}_{\text{This is cool}}$ ...too much(Sun)