Proving Identities: Need Help!

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In summary: Oh, I see what you did.You were asked to prove:$$\frac{\sin^2(x) - \cos^2(x)}{\cos^2(x)} = \tan^2(x) - 1$$on the right side, you started with:##\displaystyle \frac{\sin^2(x) - \cos^2(x)}{\cos^2(x)} \ ##and you divided the numerator and denominator by cos^2(x) to obtain:##\displaystyle \frac{\frac{\sin^2(x)}{\cos^2(x)} - \frac{\cos^2(x)}{\cos^2(x)}}{\cos^2(x)} \ ##That is not
  • #1
alexandria
169
2

Homework Statement


prove the following identity [/B]
upload_2016-8-14_2-22-59.png


Homework Equations


no equations required

The Attempt at a Solution


I've been trying to prove this identity, but no matter what I do, I can't seem to make both sides the same
here is my answer to this qts so far: can someone please tell me what I have to do next?
Im pretty sure I'm supposed to be getting tan(x) - 1 on both sides. So, should I leave the left side the same (keep it as tan(x) -1) and work only with the right side, until both sides are equal?
any help would be appreciated. Thanks.
upload_2016-8-14_2-25-13.png
Sorry for accidently posting the same thread twice :p
 

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  • #2
alexandria said:

Homework Statement


prove the following identity [/B]
View attachment 104676

Homework Equations


no equations required

The Attempt at a Solution


I've been trying to prove this identity, but no matter what I do, I can't seem to make both sides the same
here is my answer to this qts so far: can someone please tell me what I have to do next?
Im pretty sure I'm supposed to be getting tan(x) - 1 on both sides. So, should I leave the left side the same (keep it as tan(x) -1) and work only with the right side, until both sides are equal?
any help would be appreciated. Thanks.
View attachment 104677Sorry for accidently posting the same thread twice :p
When you divide a sum or difference, you have to divide all terms. ##\frac{\sin(θ)-\cos(θ)}{\cos(θ)}## is not sin(θ)-1. And you made the same error in the right side.
Leave the left side as it is, and work with the right side. Divide both numerator and denominator with cos2(θ), for example.
 
  • #3
Thanks for the quick reply!
Ok so i understand that i am supposed to leave the left side the same, but what do i do with the right side. You mentioned dividing both numerator and denominator with cos^2(θ), how do i do that?
so is it:
sin^2(x) - cos^2(x) /cos^2(x)
-------------------------------------
sin(x) cos(x) + cos^2(x) / cos^2(x)

i was originally using trigonometric identities:
for ex: sin^2(x) + cos^2(x) = 1 so i was assuming that sin^2(x) - cos^2(x) = -1
and so i replaced sin^2(x) - cos^2(x) with -1
is this correct?
 
  • #4
alexandria said:
Thanks for the quick reply!
Ok so i understand that i am supposed to leave the left side the same, but what do i do with the right side. You mentioned dividing both numerator and denominator with cos^2(θ), how do i do that?
so is it:
((sin^2(x) - cos^2(x)) /cos^2(x)
-------------------------------------
(sin(x) cos(x) + cos^2(x)) / cos^2(x)
No. You have to divide the whole expressions, not only the last terms. Do not omit the parentheses.
alexandria said:
i was originally using trigonometric identities:
for ex: sin^2(x) + cos^2(x) = 1 so i was assuming that sin^2(x) - cos^2(x) = -1
and so i replaced sin^2(x) - cos^2(x) with -1
is this correct?
No. It is wrong. sin^2(x) - cos^2(x) is not -1. Try with numbers. 2/3 + 1/3 = 1, is it true, that 2/3 - 1/3 = -1?
 
  • #5
alexandria said:
Thanks for the quick reply!
Ok so i understand that i am supposed to leave the left side the same, but what do i do with the right side. You mentioned dividing both numerator and denominator with cos^2(θ), how do i do that?
so is it:
sin^2(x) - cos^2(x) /cos^2(x)
-------------------------------------
sin(x) cos(x) + cos^2(x) / cos^2(x)

i was originally using trigonometric identities:
for ex: sin^2(x) + cos^2(x) = 1 so i was assuming that sin^2(x) - cos^2(x) = -1
and so i replaced sin^2(x) - cos^2(x) with -1
is this correct?

sin^2 x + cos^2 x = 1
=> -sin^2 x - cos^2 x = -1 and not what you wrote.

Leave the left side as it is. In the right side, we notice that we can use the formula: a^2 - b^2 = (a-b)(a+b) in the numerator. Try to find the factors of the denumerator and the problem becomes quite trivial then.
 
  • #6
ok so for the right side,if I use the formula: a^2 - b^2 = (a-b)(a+b)
then it becomes:

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

how do i solve this??
 
  • #7
You're not getting the right push so far. You need to show that a = b/c. Thus ac = b. Cross-multiply (tan theta - 1) by the denominator (sin theta cos theta + cos^2 theta) theta and simplify.
 
  • #8
srry I am very confused, what do you mean by cross multiply and why would I need to cross multiply tan(x) - 1 by the denominator of the right side?
Can you provide an example!
 
  • #9
ok so i came up with an answer, can someone tell me if I am on the right track?

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

= tan^2(x)
----------
tan(x) + 1

= tan(x) tan(x)
-------------
tan(x) + 1

= tan(x) + 1

im not sure what to do from here.
 
  • #10
alexandria said:
ok so i came up with an answer, can someone tell me if I am on the right track?

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

= tan^2(x)
----------
tan(x) + 1

The last equation is wrong.
 
  • #11
can you please elaborate?
 
  • #12
alexandria said:
ok so i came up with an answer, can someone tell me if I am on the right track?

((sin^2(x) - cos^2(x)) /cos^2(x)
-------------------------------------
(sin(x) cos(x) + cos^2(x)) / cos^2(x)
Look at that numerator again !

What is ##\displaystyle \frac{\sin^2(x) - \cos^2(x)}{\cos^2(x)} \ ## ?

It's not
= tan^2(x)

You may need to review: adding and subtracting fractions !
 
  • #13
for the numerator:

sin^2(x) - cos^2(x) / cos^2 (x)
inorder to find the answer, I have to divide cos^2(x) by each term in the expression:
so sin^2(x) / cos^2(x) = tan^2(x) (since sin(x)/cos(x) = tan(x))
and -cos^2(x) /cos^2(x) = -1

so is the answer tan^2(x) - 1
?
 
  • #14
alexandria said:
for the numerator:

sin^2(x) - cos^2(x) / cos^2 (x)
inorder to find the answer, I have to divide cos^2(x) by each term in the expression:
so sin^2(x) / cos^2(x) = tan^2(x) (since sin(x)/cos(x) = tan(x))
and -cos^2(x) /cos^2(x) = -1

so is the answer tan^2(x) - 1
?

alexandria said:
ok so for the right side,if I use the formula: a^2 - b^2 = (a-b)(a+b)
then it becomes:

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

how do i solve this??

Factor out a cosx in the denumerator:

[sin^2 (x) - cos^2(x)]/[sinxcosx + cos^2 x]
= [(sinx - cosx)(sinx + cosx)]/[cosx(sinx + cosx)]

Then use (a+b)/c = a/c + b/c
 
  • #15
alexandria said:
for the numerator:

(sin^2(x) - cos^2(x)) / cos^2 (x)
inorder to find the answer, I have to divide cos^2(x) by each term in the expression:
so sin^2(x) / cos^2(x) = tan^2(x) (since sin(x)/cos(x) = tan(x))
and -cos^2(x) /cos^2(x) = -1

so is the answer tan^2(x) - 1
?
Yes.

So make that correction in the following.
alexandria said:
ok so i came up with an answer, can someone tell me if I am on the right track?

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

= tan^2(x)
----------
tan(x) + 1

...
That gives

## \displaystyle \frac{\tan^2(x) - 1}{\tan(x)+1} \ ##
 
  • #16
((sin^2(x) - cos^2(x)) /cos^2(x)
-------------------------------------
(sin(x) cos(x) + cos^2(x)) / cos^2(x)= tan^2(x) - 1
--------------
tan(x) + 1

= tan(x) - 1

is this correct?
 
  • #17
alexandria said:
((sin^2(x) - cos^2(x)) /cos^2(x)
-------------------------------------
(sin(x) cos(x) + cos^2(x)) / cos^2(x)= tan^2(x) - 1
--------------
tan(x) + 1

= tan(x) - 1

is this correct?
That is correct.

Can you justify that last step ?
 
  • #18
ok so,

tan^2(x) - 1
---------------
tan(x) + 1

= tan(x) tan(x) - 1
------------------ ( I am pretty sure tan(x) crosses out at this part)
tan(x) + 1

= tan(x) - 1 / 1

= tan(x) - 1

is this process correct?
 
  • #19
alexandria said:
ok so,

tan^2(x) - 1
---------------
tan(x) + 1

= tan(x) tan(x) - 1
------------------ ( I am pretty sure tan(x) crosses out at this part)
tan(x) + 1

= tan(x) - 1 / 1

= tan(x) - 1

is this process correct?
That's what I suspected.

That is not the correct way to obtain the result.

a2 -1 factors to (a - 1)(a + 1)..

So, can you factor ##\ \tan^2(x) - 1 \ ?##
 
  • #20
ok so,

(tan(x) - 1) (tan(x) + 1) (cross out the expression tan(x) + 1)
-----------------------------
tan(x) + 1

= tan(x) - 1
 
  • #21
alexandria said:
ok so,

(tan(x) - 1) (tan(x) + 1) (cross out the expression tan(x) + 1)
-----------------------------
tan(x) + 1

= tan(x) - 1
That's more like it !
 
  • #22
thanks so much for all the help, I really appreciate it! :smile:
 
  • #23
Well take RHS.
cos θ is obviously a factor of the denominator. So factorise it.
Now look at the numerator. This is the difference of two squares. Know how you can factorise - OK how you can write difference of two squares? (a2 - b2)?
Do that then you see one of the factors is the same as the factor of the denominator. Then it all falls out easily.

Why can I do that in my head, and you have spent an hour on it? Well partly it's because I don't have to do it. So I'm relaxed about it You might need to relax
Secondly I'm using a Polya 'How To Solve It' principle - asking self "have you seen anything like this before?".
Sure you see trig the formula and you start thinking of other trig formulae – you might not recognise immediately the resemblance with (a2 - b2). That may be where being relaxed helps
 
  • #24

Homework Statement


Prove the identity:
upload_2016-8-16_20-31-54.png


Homework Equations


upload_2016-8-16_20-32-30.png


The Attempt at a Solution


upload_2016-8-16_20-31-4.png


Can someone tell me if this is correct? I am not sure if I should have worked on both the right and left side until they were the same, instead of just the left side?
 
  • #25
Evangeline101 said:

Homework Statement


Prove the identity:
View attachment 104794

Homework Equations


View attachment 104795

The Attempt at a Solution


View attachment 104793

Can someone tell me if this is correct? I am not sure if I should have worked on both the right and left side until they were the same, instead of just the left side?

Looks okay,

The second step on the left hand side is a bit strange, but then you fixed it the next step, so probably just a typo.
 
  • #26
Student100 said:
The second step on the left hand side is a bit strange, but then you fixed it the next step, so probably just a typo.

This step?
upload_2016-8-16_20-47-4.png

should I just take it out then?
 
  • #27
Evangeline101 said:
This step?
View attachment 104796
should I just take it out then?

The next step, it's not equal to ##\frac{\sin{x}+\sin{x}}{\cos{x}}##
 
  • #28
It seems to me that working the right hand side would be easier than working the left hand side.
 
  • #29
Student100 said:
The next step, it's not equal to sinx+sinxcosx

Okay, I edited it:

upload_2016-8-16_21-42-25.png

upload_2016-8-16_21-43-1.png


Is this better?
 
  • #31
Student100 said:
That's fine. You just need to be careful, the = is saying something very precise.

what do you mean?
 
  • #32
Evangeline101 said:
what do you mean?

The above line that I posted, you're saying the original expression is equal to 2tan(x), which isn't true. I know it was a typo, just make sure you're careful is all. No sense losing points on a typo. =)

The fixed solution looks fine to me.
 
  • #33
Student100 said:
The above line that I posted, you're saying the original expression is equal to 2tan(x), which isn't true. I know it was a typo, just make sure you're careful is all.

Okay, thanks for the help! :)
 
  • #34
Sorry for responding to this thread after it has been marked solved.

Ok, I am proving the exact same identity, I looked through the thread and the explanations were very helpful.

I wrote out the answer, and would just like to know if I have written the answer properly:

LS= tan x - 1
upload_2016-8-16_23-29-36.png
 
Last edited:
  • #35
Yes I think that's right now.
Hope you see it as simple now, and what you were missing.

One of the handful of formulae constantly useful to remember fluently, difference of two squares - and then difference of two n-th powers.
 

What is the purpose of proving identities?

The purpose of proving identities is to show that two mathematical expressions are equivalent. This is important in solving equations and simplifying expressions.

How do you prove identities?

There are various methods for proving identities, including using algebraic manipulations, substitution, and using known trigonometric identities. It is also helpful to have a solid understanding of mathematical properties and rules.

Why is it important to show all steps when proving identities?

Showing all steps when proving identities is important because it allows for a clear understanding of the thought process and logic behind the proof. It also allows for easier identification of any errors made in the process.

What are some common mistakes to avoid when proving identities?

Some common mistakes to avoid when proving identities include incorrect algebraic manipulations, forgetting to use parentheses, and not simplifying expressions fully. It is also important to check for extraneous solutions and to double check all steps.

How can I improve my skills in proving identities?

To improve your skills in proving identities, it is important to practice regularly and to understand the underlying concepts and rules. You can also seek help from a tutor or teacher, and use online resources and practice problems to strengthen your skills.

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