- #1

acen_gr

- 63

- 0

Is this identity possible?

[tex]cot 2x = \frac{cos3x + cosx}{sin 3x + sinx}[/tex]

Thanks!

[tex]cot 2x = \frac{cos3x + cosx}{sin 3x + sinx}[/tex]

Thanks!

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter acen_gr
- Start date

- #1

acen_gr

- 63

- 0

Is this identity possible?

[tex]cot 2x = \frac{cos3x + cosx}{sin 3x + sinx}[/tex]

Thanks!

[tex]cot 2x = \frac{cos3x + cosx}{sin 3x + sinx}[/tex]

Thanks!

- #2

Mark44

Mentor

- 36,661

- 8,659

Is this homework?

- #3

acen_gr

- 63

- 0

Is this homework?

Not homework. I'm just asking if it's possible for the expressions on both sides to be identical or the equation is an identity or not. Because I've tried to work on it but I couldn't make them identical.

- #4

HallsofIvy

Science Advisor

Homework Helper

- 43,021

- 971

- #5

uart

Science Advisor

- 2,797

- 21

Is this identity possible?

[tex]cot 2x = \frac{cos3x + cosx}{sin 3x + sinx}[/tex]

Thanks!

Try cross multiplying to get:

[tex]\cos 2x (\sin 3x + \sin x) = \sin 2x (\cos 3x + \cos x)[/tex]

and then apply the half sum/difference identities to all of the sin cos products on each side of the equation. It comes out fairly easy with this step.

Use:

[tex]\sin a \cos b = \frac{1}{2} \left[ \sin(a+b) + \sin(a-b) \right][/tex]

- #6

mathman

Science Advisor

- 8,084

- 550

For example: cot(2x) = i(u

Do the similar steps for the right hand side, clear the denominators and end up with an identity.

- #7

acen_gr

- 63

- 0

Try cross multiplying to get:

[tex]\cos 2x (\sin 3x + \sin x) = \sin 2x (\cos 3x + \cos x)[/tex]

and then apply the half sum/difference identities to all of the sin cos products on each side of the equation. It comes out fairly easy with this step.

Use:

[tex]\sin a \cos b = \frac{1}{2} \left[ \sin(a+b) + \sin(a-b) \right][/tex]

[tex]\cos 2x (\sin 3x + \sin x) = \sin 2x (\cos 3x + \cos x)[/tex]

[tex]\cos 2x \sin3x + \cos2x sinx = \sin2x \cos3x + \sin2x cosx[/tex]

Should I end here? I think both doesn't equal up. Or should I go further by extracting cos2x, sin2x, cos3x, and sin3x ?

- #8

mathman

Science Advisor

- 8,084

- 550

[tex]\cos 2x (\sin 3x + \sin x) = \sin 2x (\cos 3x + \cos x)[/tex]

[tex]\cos 2x \sin3x + \cos2x sinx = \sin2x \cos3x + \sin2x cosx[/tex]

Should I end here? I think both doesn't equal up. Or should I go further by extracting cos2x, sin2x, cos3x, and sin3x ?

You need to end up with sinx and cosx only, using identities for cos2x, etc. I think my method (using representation in e

- #9

uart

Science Advisor

- 2,797

- 21

No, apply the half sum-difference formula (that I gave above) to each of the sin-cos products.[tex]\cos 2x (\sin 3x + \sin x) = \sin 2x (\cos 3x + \cos x)[/tex]

[tex]\cos 2x \sin3x + \cos2x sinx = \sin2x \cos3x + \sin2x cosx[/tex]

Should I end here?

Share:

- Last Post

- Replies
- 4

- Views
- 383

- Last Post

- Replies
- 11

- Views
- 607

- Replies
- 3

- Views
- 767

- Replies
- 3

- Views
- 854

- Last Post

- Replies
- 1

- Views
- 524

- Last Post

- Replies
- 2

- Views
- 320

- Last Post

- Replies
- 2

- Views
- 167

- Last Post

- Replies
- 8

- Views
- 488

- Last Post

- Replies
- 22

- Views
- 922

- Last Post

- Replies
- 0

- Views
- 398