# Proving that an equation is an identity

1. May 1, 2006

### Aya

Hi, i need help with this question

...now what

2. May 1, 2006

### GregA

You're trying to prove that $$1 +cot^2\theta$$ is an identity for $$sin^2\theta$$? The maximum value of the RHS is 1...the max possible value of the LHS isn't!...further more, 1 is the minimum value of the LHS.

Perhaps they meant for the RHS: $$csc^2\theta$$...this can be proved rather quickly from the pythagorean identity,

Last edited: May 1, 2006
3. May 1, 2006

### Aya

I am trying to prove that

1+ 1/tan2x = sin2x

4. May 1, 2006

### GregA

In this case, the question you are asked to prove is ridiculous... If we let x = a small angle, tan2x's reciprocal will still by far exceed sin2x..

If someone *can* prove this I shall be very surprised

Last edited: May 1, 2006
5. May 1, 2006

### Staff: Mentor

tan^2x, tan2x....

Anyway, I googled trigonometric identities for you, and here's the first hit. Do any of these look like what you are trying to prove?

http://aleph0.clarku.edu/~djoyce/java/trig/identities.html

6. May 1, 2006

### VietDao29

Is that sin2x or 1 / sin2x?
If that's sin2x, then you don't need to prove it anymore, the identity is wrong.
If that's 1 / sin2x, then you can continue by using the Pythagorean Identity: sin2x + cos2x = 1.
So:
$$1 + \frac{\cos ^ 2 x}{\sin ^ 2 x} = \frac{\sin ^ 2 x}{\sin ^ 2 x} + \frac{\cos ^ 2 x}{\sin ^ 2 x} = \frac{\sin ^ 2 x + \cos ^ 2 x}{\sin ^ 2 x} = ...$$.
I'll leave you finish the rest... :)
By the way, does the problem ask you to prove the identity or to solve the equation?

Last edited: May 2, 2006
7. May 1, 2006

### Curious3141

Scan the question and post it please.

8. May 2, 2006

### Aya

It says solve but the teacher said that all the questions that were assigned are possible to prove

Sorry, i don't have a scaner

Thanks for everyones help

9. May 2, 2006

### Aya

I have another question

4/cos^2x -5 = 4 tan ^2 x-1

(sin x + cos x )2 = 1 + 2sin x cos x

pleas help with these, I have a mth test tomarow

And I am curious to know, why is the section called pre-calculus math? (besides the obvious reason) is that what the coarse is called in the US?

Last edited: May 2, 2006
10. May 2, 2006

### VietDao29

There's one big diference between solve the equation, and prove the identity.
When you are ask to prove the identity, the equation is always correct, i.e, no matter what the input is, the LHS is always equal the RHS. However, when the equation needs to be solved, only a few inputs will satisfy the equation, sometime no input will satisfy it. That's the difference between prove, and solve.
-------------------
Take your first problem as the example:
Solve the equation:
$$1 + \frac{1}{\tan ^ 2 x} = \sin ^ 2 x$$
$$\Leftrightarrow \frac{1}{\sin ^ 2 x} = \sin ^ 2 x$$
$$\Leftrightarrow 1 = \sin ^ 4 x$$
$$\Leftrightarrow \left[ \begin{array}{l} \sin(x) = 1 \\ \sin(x) = -1 \end{array} \right.$$
$$\Leftrightarrow x = \frac{\pi}{2} + k \pi, \ k \in \mathbb{Z}$$
i.e, if your inputs are of the form: $$\frac{\pi}{2} + k \pi , \ k \in \mathbb{Z}$$, then the LHS will equal the RHS, if it's not, then the LHS is not equal the RHS.
Can you get this? :)
---------------------
Have you done anything? To learn maths is to do maths. At least, you should give those problem a try before you post it here, right?
The second identity can be proven using the Pythagorean Identity, and expanding all the terms out.

Last edited: May 2, 2006
11. May 2, 2006

### Aya

^ in the second step

1/tan^2x
1/sin^2x

where did the cos^2x go?

12. May 2, 2006

### VietDao29

The LHS is:
$$1 + \frac{1}{\tan ^ 2 x}$$, not just tan2x.
$$1 + \frac{1}{\tan ^ 2 x} = 1 + \frac{\cos ^ 2 x}{\sin ^ 2 x} = \frac{\sin ^ 2 x + \cos ^ 2 x}{\sin ^ 2 x} = \frac{1}{\sin ^ 2 x}$$.
Can you get this now? :)
Have you tried the 2nd, and the 3rd problem? :)

13. May 2, 2006

### Aya

^ Ok thank you
sin x + cos x )2 = 1 + 2sin x cos x

R.H.S.

sin x + cos x )2

sin2 x + 2 sin x cos x + cos2 x

(sin2 x + cos2 x) + 2 sin x cos x

1+ 2 sin x cos x

L.H.S

1 + 2sin x cos x

yes I got the third one

4/cos^2x -5 = 4 tan ^2 x-1

L.H.S.

4 sin^2X /cos ^2x -1

Find a common denominatior
4cos^2 ( sin ^2x)- cos ^2/cos^2x

??? I don't know what to do next, are the abouve steps correct?

14. May 2, 2006

### VietDao29

Yeah, this is correct. :)

You can stop right here, what's $$\frac{\sin ^ 2 x}{\cos ^ 2 x}$$ equal to? :)

15. May 2, 2006

### Aya

Tan^2 x, but I'm tring to prove that 4 tan ^2 x-1 is equal to 4/cos^2x -5. If i change it back to Tan^2 x, won't it be the same as the orriginal?

16. May 2, 2006

### VietDao29

Ah, sorry, I thought you were going from the LHS to the RHS.
$$\frac{4 \sin ^ 2 x - \cos ^ 2 x}{\cos ^ 2 x}$$. Now you must do something make the numerator have the 4 to obtain $$\frac{4}{\cos ^ 2 x}$$, right?
So:
$$\frac{4 \sin ^ 2 x - \cos ^ 2 x}{\cos ^ 2 x} = \frac{4 + 4 \sin ^ 2 x - 4 -\cos ^ 2 x}{\cos ^ 2 x} = ...$$
Can you go from here? Hint : Use the Pythagorean Identity.
However going from the LHS is easier:
$$\frac{4}{\cos ^ 2 x} - 5$$, now the RHS has '-1', so you just keep -1 there.
$$\frac{4}{\cos ^ 2 x} - 5 = \frac{4}{\cos ^ 2 x} - 4 - 1 = \frac{4 - 4 \cos ^ 2 x}{\cos ^ 2 x} - 1 = ...$$.
Can you go from here? :)

17. May 7, 2006

### Aya

^yes, thanks. I already had the test and I got all the proving identities questions correct! Thanks for everyone’s help!