# Simplification of Trig Functions TWO

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• Sep 8th 2006, 04:21 AM
classicstrings
Simplification 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 AM
malaygoel
Quote:

Originally Posted by classicstrings
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!

simply divide
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 AM
classicstrings
So,

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

Right?
• Sep 8th 2006, 04:27 AM
malaygoel
Quote:

Originally Posted by classicstrings
So,

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

Right?

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

Keep Smiling
Malay
• Sep 8th 2006, 04:30 AM
classicstrings
Done thanks!
• Sep 8th 2006, 07:31 AM
ThePerfectHacker
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
you could divide only if a in not equal to b

You mean $\displaystyle a\not = -b$
• Sep 8th 2006, 07:47 PM
classicstrings
Ah yes factoring, thanks for the 2nd opinion!
• Sep 8th 2006, 10:33 PM
CaptainBlack
Quote:

Originally Posted by classicstrings
Ah yes factoring, thanks for the 2nd opinion!

The interesting thing here is that Malay would not have suggested that
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