Hi, I have

[(sin^2theta) - (cos^2theta) ]/ (sin(theta) + cos(theta))

And I need to simplify it to sin(theta) - cos(theta).

How do I do that? Thanks!

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- Sep 8th 2006, 04:21 AMclassicstringsSimplification of Trig Functions TWO
Hi, I have

[(sin^2theta) - (cos^2theta) ]/ (sin(theta) + cos(theta))

And I need to simplify it to sin(theta) - cos(theta).

How do I do that? Thanks! - Sep 8th 2006, 04:24 AMmalaygoelQuote:

Originally Posted by**classicstrings**

for siplification, use sin(theta)=a;cos(theta)=b

you could divide only if a in not equal to b

Keep Smiling

Malay - Sep 8th 2006, 04:26 AMclassicstrings
So,

(a^2 - b^2)/ (a+b)

Right? - Sep 8th 2006, 04:27 AMmalaygoelQuote:

Originally Posted by**classicstrings**

you needn't use any trigonometric formulae here. It is simple algebraic division

Keep Smiling

Malay - Sep 8th 2006, 04:30 AMclassicstrings
Done thanks!

- Sep 8th 2006, 07:31 AMThePerfectHacker
You can also factor, instead of long divsion.

$\displaystyle \frac{(\sin x+\cos x)(\sin x-\cos x)}{\sin x+\cos x}$

Quote:

Originally Posted by**malaygoel**

- Sep 8th 2006, 07:47 PMclassicstrings
Ah yes factoring, thanks for the 2nd opinion!

- Sep 8th 2006, 10:33 PMCaptainBlackQuote:

Originally Posted by**classicstrings**

you simply divide if he had not already recognised that the numerator

factored from the difference of two squares form that it has and that

one of these factors was the denominator.

(At least that is my take on what was going on)

RonL