Prove the following equasion:

**Punkbuster** · December 16th 2010, 04:56 AM

Hey guys,
I have a question to prove the following equation:
$Sin^2aCos^2B - Sin^2Bcos^2a = Cos^2B - Cos^2a<br />$
What I mean by 'proving' is that you have to take one side of the equation and reach the other side...
Usually its easier to take the left side.
B = Beta a = Alpha
Cheers,
Punkbuster

**tonio** · December 16th 2010, 05:22 AM

Originally Posted by Punkbuster

Hey guys,
I have a question to prove the following equation:
$Sin^2aCos^2B - Sin^2Bcos^2a = Cos^2B - Cos^2a<br />$
What I mean by 'proving' is that you have to take one side of the equation and reach the other side...
Usually its easier to take the left side.
B = Beta a = Alpha
Cheers,
Punkbuster

Dirty algebra trick + Algebra + trigonometric Pythagoras theorem. Read carefully the following and justify each step:

$\displaystyle{\sin^2\alpha\cos^2\beta-\sin^2\beta\cos^2\alpha=\sin^2\alpha\cos^2\beta-\sin^2\alpha\cos^2\alpha+\sin^2\alpha\cos^2\alpha-\sin^2\beta\cos^2\alpha=}$

$=\displaystyle{\sin^2\alpha(\cos^2\beta-\cos^2\alpha)+\cos^2\alpha(\sin^2\alpha-\sin^2\beta)=\sin^2\alpha(cos^2\beta-\cos^2\alpha)+\cos^2\alpha(1-\cos^2\alpha-(1-\cos^2\beta))=}$

$=\displaystyle{\sin^2\alpha(\cos^2\beta-\cos^2\alpha)+\cos^2\alpha(\cos^2\beta-\cos^2\alpha)}=$ ....end the proof.

Tonio

**Punkbuster** · December 16th 2010, 05:25 AM

Originally Posted by tonio

Dirty algebra trick + Algebra + trigonometric Pythagoras theorem. Read carefully the following and justify each step:

$\displaystyle{\sin^2\alpha\cos^2\beta-\sin^2\beta\cos^2\alpha=\sin^2\alpha\cos^2\beta-\sin^2\alpha\cos^2\alpha+\sin^2\alpha\cos^2\alpha-\sin^2\beta\cos^2\alpha=}$

$=\displaystyle{\sin^2\alpha(\cos^2\beta-\cos^2\alpha)+\cos^2\alpha(\sin^2\alpha-\sin^2\beta)=\sin^2\alpha(cos^2\beta-\cos^2\alpha)+\cos^2\alpha(1-\cos^2\alpha-(1-\cos^2\beta))=}$

$=\displaystyle{\sin^2\alpha(\cos^2\beta-\cos^2\alpha)+\cos^2\alpha(\cos^2\beta-\cos^2\alpha)}=$ ....end the proof.

Tonio

Damn!! Can you believe I might need to do something like that in an exam??

**Soroban** · December 16th 2010, 05:25 AM

Hello, Punkbuster!

Use the identity: . $\sin^2\!\theta + \cos^2\!\theta \:=\:1 \quad\Rightarrow\quad \sin^2\!\theta \:=\:1-\cos^2\!\theta$

$\text{Prove: }\:\sin^2\!A\cos^2\!B - \sin^2\!B\cos^2\!A \:=\:\cos^2\!B - \cos^2\!A$

$\sin^2\!A\cos^2\!B - \sin^2\!B\cos^2\!A$

. . $=\;(1-\cos^2\!A)\cos^2\!B - (1-\cos^2\!B)\cos^2\!A$

. . $=\;\cos^2\!B - \cos^2\!A\cos^2\!B - \cos^2\!A + \cos^2\!A\cos^2\!B$

. . $=\;\cos^2\!B - \cos^2\!A$

**Punkbuster** · December 16th 2010, 05:37 AM

Originally Posted by Soroban

Hello, Punkbuster!

Use the identity: . $\sin^2\!\theta + \cos^2\!\theta \:=\:1 \quad\Rightarrow\quad \sin^2\!\theta \:=\:1-\cos^2\!\theta$

$\sin^2\!A\cos^2\!B - \sin^2\!B\cos^2\!A$

. . $=\;(1-\cos^2\!A)\cos^2\!B - (1-\cos^2\!B)\cos^2\!A$

. . $=\;\cos^2\!B - \cos^2\!A\cos^2\!B - \cos^2\!A + \cos^2\!A\cos^2\!B$

. . $=\;\cos^2\!B - \cos^2\!A$

I was about to come here and repost what you said...
But then I found that you raced me
Thanks

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