# Homework Help: Proving identities

1. Aug 14, 2016

### alexandria

1. The problem statement, all variables and given/known data
prove the following identity

2. Relevant equations
no equations required
3. 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.

Sorry for accidently posting the same thread twice :p

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2. Aug 14, 2016

### ehild

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. Aug 14, 2016

### alexandria

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. Aug 14, 2016

### ehild

No. You have to divide the whole expressions, not only the last terms. Do not omit the parentheses.
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. Aug 14, 2016

### Math_QED

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. Aug 14, 2016

### alexandria

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. Aug 14, 2016

### dextercioby

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. Aug 14, 2016

### alexandria

srry im 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. Aug 14, 2016

### alexandria

ok so i came up with an answer, can someone tell me if im 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. Aug 14, 2016

### ehild

The last equation is wrong.

11. Aug 14, 2016

### alexandria

12. Aug 15, 2016

### SammyS

Staff Emeritus
Look at that numerator again !

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

It's not
You may need to review: adding and subtracting fractions !

13. Aug 15, 2016

### alexandria

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. Aug 15, 2016

### Math_QED

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. Aug 15, 2016

### SammyS

Staff Emeritus
Yes.

So make that correction in the following.
That gives

$\displaystyle \frac{\tan^2(x) - 1}{\tan(x)+1} \$

16. Aug 15, 2016

### alexandria

((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. Aug 15, 2016

### SammyS

Staff Emeritus
That is correct.

Can you justify that last step ?

18. Aug 15, 2016

### alexandria

ok so,

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

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

= tan(x) - 1 / 1

= tan(x) - 1

is this process correct?

19. Aug 16, 2016

### SammyS

Staff Emeritus
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. Aug 16, 2016

### alexandria

ok so,

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

= tan(x) - 1

21. Aug 16, 2016

### SammyS

Staff Emeritus
That's more like it !

22. Aug 16, 2016

### alexandria

thanks so much for all the help, I really appreciate it!

23. Aug 16, 2016

### epenguin

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. Aug 16, 2016

### Evangeline101

1. The problem statement, all variables and given/known data
Prove the identity:

2. Relevant equations

3. The attempt at a solution

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. Aug 16, 2016

### Student100

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.