couple identity problems

corsec

New member
Joined
Apr 19, 2006
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13
hello I'm needing to simplify these two problems

cos^4(x)-sin^4(x)
--------------------
cos^2(x)-sin^2(x)

using pythagareom identity I've deduced the denominator should be

cos^2(x) -1 = -sin^2(x)

but as for the top i'm lost

looking at my choices of ansers I'm guessing it should be 2Cos^2(x) -1 since the top is twice the exponent as the bottom but i'm not sure.

the other one is

csc(x)
------------
tan x + cot x

I'm not even sure what identitity to look for in this one any ideas would be great!
Thank you.
 
If I'm reading it correctly, the numerator in the top problem looks like the difference of 2 squares so it would probably be easiest to seperate it into (cos^2(x)+sin^2(x))*(cos^2(x)-sin^2(x)) and just simplify from there. For the bottom, I would suggest changing the csc(X) to 1/sin(x) and change the cot(x) and tan(x) into terms of sin(x) and cos(x) and multiply everything out and see where that leaves you.
 
Hello, corsec!

rzornick's suggestions are the best . . .

\(\displaystyle \L\frac{\cos^4x\,-\,\sin^4x}{\cos^2x\,-\,\sin^2x}\)
Factor: \(\displaystyle \L\,\frac{(\sout{\cos^2x\,-\,\sin^2x})(\cos^2x\,+\,\sin^2x)}{\sout{\cos^2x\,-\,\sin^2x}}\;= \;\cos^2x\,+\,\sin^2x\;=\;1\)


\(\displaystyle \L\frac{\csc x}{\tan x\,+\,\cot x}\)
We have: \(\displaystyle \L\,\frac{\frac{1}{\sin x}}{\frac{\sin x}{\cos x}\,+\,\frac{\cos x}{\sin x}}\)

Multiply top and bottom by \(\displaystyle \sin x\cos x:\L\;\;\frac{\sin x\cos x\left(\frac{1}{\sin x}\right)}{\sin x\cos x\left(\frac{\sin x}{\cos x}\,+\,\frac{\cos x}{\sin x}\right)}\)

\(\displaystyle \L\;\;\;= \;\frac{\cos x}{\sin^2x\,+\,\cos^2x} \;= \;\frac{\cos x}{1} \;= \;\cos x\)
 
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