proving identity: (Cos^4x-Sin^4x)/1-Tan^4x = Cos^4x

[peter0516]

ok well i was assigned this math problem and i have no clue how to do this. we need to prove that this identity is correct.

(Cos^4x-Sin^4x)/1-Tan^4x = Cos^4x

if anyone can explain this please help me thank you

 

[Deleted member 4993]

Re: proving identites to be correct! please look

ok well i was assigned this math problem and i have no clue how to do this. we need to prove that this identity is correct.

(Cos^4x-Sin^4x)/1-Tan^4x = Cos^4x

if anyone can explain this please help me thank you

use

$\displaystyle Tan(x) = \frac{Sin(x)}{Cos(x)}$

 

[peter0516]

Re: proving identites to be correct! please look

that wont work since its raised to the fourth power

 

[Deleted member 4993]

Re: proving identites to be correct! please look

that wont work since its raised to the fourth power

what do you mean?

$\displaystyle Tan^4(x) = [Tan(x)]^4$

 

[peter0516]

Re: proving identites to be correct! please look

idk but how would i prove it so that one side equals the other side

 

[Deleted member 4993]

Re: proving identites to be correct! please look

Like I told - start replacing - show work - tell me where you are stuck.

You are waving white flag before you start....

 

[peter0516]

Re: proving identites to be correct! please look

i got down to (Cos^2x)/1-(Sin^4x/Cos^4x) - (Sin^2x)/1-(Sin^4x/Cos^4x) = cos^4 x

 

[Deleted member 4993]

Re: proving identites to be correct! please look

i got down to (Cos^2x)/1-(Sin^4x/Cos^4x) - (Sin^2x)/1-(Sin^4x/Cos^4x) = cos^4 x

deal with the denominator first:

$\displaystyle 1 -Tan^4(x) = 1 - \frac{Sin^4(x)}{Cos^4(x)} \, = \, \frac{Cos^4(x) - Sin^4(x)}{Cos^4(x)}$....edited

Now continue....

 

[peter0516]

Re: proving identites to be correct! please look

that dosent make sense becuase tanx= sin x/cos x

 

[o_O]

Re: proving identites to be correct! please look

$\displaystyle \frac{cos^{4}x - sin^{4}x}{1 - tan^{4}x}$

$\displaystyle = \frac{cos^{4}x - sin^{4}x}{1 - \frac{sin^{4}x}{cos^{4}x}}$

What he was trying to say is to combine the denominator into one single fraction. Something is sure to cancel.

A mere typo was made but the idea is there.

 

[peter0516]

Re: proving identites to be correct! please look

oo ok
now i got down to
(-Sin^4x)/ cos^4x = cos^4 x which is tan^4x but that dosent equal cos ^4 x

 

[o_O]

Re: proving identites to be correct! please look

How did you arrive at that answer?

 

[peter0516]

Re: proving identites to be correct! please look

because i did Cos^4(X) - Sin^4(X) / Cos^4(X) then i crossed out the cosine and got it to sin over cos.

 

[o_O]

Re: proving identites to be correct! please look

Ok so it looks like you combined the denominator correctly. So it should be:

$\displaystyle \frac{cos^{4}x - sin^{4}x}{1 - \frac{sin^{4}x}{cos^{4}x}}$

$\displaystyle =\frac{cos^{4}x - sin^{4}x}{\frac{cos^{4}x}{cos^{4}x} - \frac{sin^{4}x}{cos{4}x}}$

$\displaystyle =\frac{cos^{4}x - sin^{4}x}{\frac{cos^{4}x - sin^{4}x}{cos^{4}x}}$

You cannot just cancel cos[sup:2iz72ens]4[/sup:2iz72ens]x without taking into consideration of sin[sup:2iz72ens]4[/sup:2iz72ens]x.

\(\displaystyle \mbox{Denominator: } \frac{cos^{4}x - sin^{4}x}{cos^{4}x} \neq \frac{-sin^{4}x}{cos^{4}x}\)

You would have to divide each term by cos[sup:2iz72ens]4[/sup:2iz72ens]x to work but you would end up with what you started: 1 - (sinx/cosx)[sup:2iz72ens]4[/sup:2iz72ens].

If you look at your entire fraction closely, do you see anything that would immediately cancel?

 

[peter0516]

Re: proving identites to be correct! please look

oo wait cant you flip the denominator then cancel out the cos^4x and sin^4x

 

[o_O]

Re: proving identites to be correct! please look

And you're left with ... ?

 

[peter0516]

Re: proving identites to be correct! please look

cos4x =]

thank you for the help i appricate it

 

[Akshay.live]

Multiply numerator and denominator by cos^4x... Denominator reduces to cos^4x - sin^4x as cos^4x*tan^4x = sin^4x ... So cancell out numerator and denominator to leav behind the answer.