Is this Trig Identity really an identity?

[kuahji]

The textbook states something along the lines as prove the identity.

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

If you want you an work this out algebraically relatively easily to get cos x = cos x. But what if you put pi back into the original equation? You get 1 - undefined = cos x. So I graphed the two & the graphs looked the same, except the left hand side had a hole at odd multiples of pi. I showed the professor & asked him if it was still an identity. He looked in the solutions manual & its marked an identity & said he'd have to look into it further. Its been a couple of days & was just curious, it just doesn't seem to be a true a identity. What do other people think?

 

[Integral]

My guess at the way to "fix" this problem is to examin the left and right hand limits at the "holes" you should find that the limits are equal, therefore you can "fill" tbe hole with the limit value.

 

[HallsofIvy]

The textbook states something along the lines as prove the identity.

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

If you want you an work this out algebraically relatively easily to get cos x = cos x. But what if you put pi back into the original equation? You get 1 - undefined = cos x.

No, you don't. The right hand side can't be cos x, because [itex]x= \pi]. What you get is "1- undefined= -1"! Of course, it is NOT true that 1- ((sin^2 2x)/(1+ cos x))= cos x. Because the left side is not defined for $x= \pi$ while the right hand side is defined. Exactly WHAT were you supposed to prove? The "identity" you give is not correct.<br /> <br /> <blockquote data-attributes="" data-quote="" data-source="" class="bbCodeBlock bbCodeBlock--expandable bbCodeBlock--quote js-expandWatch"> <div class="bbCodeBlock-content"> <div class="bbCodeBlock-expandContent js-expandContent "> So I graphed the two & the graphs looked the same, except the left hand side had a hole at odd multiples of pi. I showed the professor & asked him if it was still an identity. He looked in the solutions manual & its marked an identity & said he'd have to look into it further. Its been a couple of days & was just curious, it just doesn't seem to be a true a identity. What do other people think? </div> </div> </blockquote>[/itex]

 

[kuahji]

The textbook says "verify each identity." In the solutions manual the problem is shown as a true identity, & the left side does break down to cos x.

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

Step 1
(1 + cos x - sin^2 x)/(1 + cos x)

Step 2
(cos x + cos^2 x)/(1 + cos x)

Step 3
cos x(1 + cos x)/(1 + cos x)

cos x = cos x

All I was really asking is if it was still an identity because the professor said he wasn't sure. The solutions manual breaks it down just like that (w/a few more steps shown). Does that make sense?

 

[Signifier]

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

Nonetheless, it can't be an "identity" because there are values of x for which the right hand side is defined but the left hand side is not (in its original form).

 

[Gib Z]

What do you define to be an "identity"? That is not clear. Algebraically, the 2 sides equal for all values of x in which they are defined, which is very reasonable. Just as 1/x = 1/x for every x in which they are defined.

 

[Dick]

I use x/x=1 all the time. Which is a rather similar sort of identity. Doesn't excuse you from checking the case x=0 though.

 

[HallsofIvy]

What do you define to be an "identity"? That is not clear. Algebraically, the 2 sides equal for all values of x in which they are defined, which is very reasonable. Just as 1/x = 1/x for every x in which they are defined.

No, they are NOT "equal for all values of x in which they are defined". The right hand side is defined for $x= \pi$ while the left side is not. They are equal for all values for which BOTH sides are defined but that is not enough to make the two functions equal. What is true is
$$1- \frac{sin x}{1+cos x}= cos x$$
for all x except odd multiples of $\pi$.

That is the same as saying that
$$\frac{x^2- 1}{x- 1}= x+ 1$$
for all x except x= 1. The two sides are different functions because they have different domains and so are not equal.