Verifying an identity

[loolmath]

I'm having a little trouble trying to verify this problem

6 sin^3 (x) cos(x) cos^2(x) − sin^2 (x) + 6 sin(x) cos^3 (x) cos^2(x) − sin^2 (x) = 3 tan(2x)

I kind of understand how to get to 3tan by factoring out the 6 in the top and rewriting the denominator as cos2x but I'm lost on where the 2x in 3tan is coming from

 

[Otis]

6 sin^3 (x) cos(x) cos^2(x) − sin^2 (x) + 6 sin(x) cos^3 (x) cos^2(x) − sin^2 (x)

… the 6 in the top and rewriting the denominator …

Hi loolmath. I don't see any ratio(s). Please type grouping symbols around the numerator and around the denominator. You may type a forward slash between them, to show the division.

Thanks

 

[loolmath]

Hi loolmath. I don't see any ratio(s). Please type grouping symbols around the numerator and around the denominator. You may type a forward slash between them, to show the division.

Thanks

Oops like this?

6 sin^3 (x) cos(x) / cos^2(x) − sin^2 (x) + 6 sin(x) cos^3 (x) / cos^2(x) − sin^2 (x) = 3 tan(2x)

 

[topsquark]

Oops like this?

6 sin^3 (x) cos(x) / cos^2(x) − sin^2 (x) + 6 sin(x) cos^3 (x) / cos^2(x) − sin^2 (x) = 3 tan(2x)

Or is it, oops, like this?
6 sin^3 (x) cos(x) / (cos^2(x) − sin^2 (x)) + 6 sin(x) cos^3 (x) / (cos^2(x) − sin^2 (x)) = 3 tan(2x)

Parentheses are important!

-Dan

 

[Otis]

Oops like this?

Nope. You didn't type grouping symbols around the numerators or denominators. Your latest version may be interpreted more than 10 different ways.

Please post a picture of the given expression, so that we may see what you're working on. Thank you.

 

[loolmath]

Nope. You didn't type grouping symbols around the numerators or denominators. Your latest version may be interpreted more than 10 different ways.

Please post a picture of the given expression, so that we may see what you're working on. Thank you.

 

[Deleted member 4993]

In the LeftHandSide, the denominators of the "fractional" expressions are same.

Moreover, those are same as the denominator of the expression at the RHS.

continue......

 

[Otis]

6 sin^3 (x) cos(x) / (cos^2(x) − sin^2 (x)) + 6 sin(x) cos^3 (x) / (cos^2(x) − sin^2 (x)) = 3 tan(2x)

Yes, that's equivalent to loolmath's image, which I'll show as:

[6*sin^3(x)*cos(x)] / [cos^2(x) − sin^2(x)] + [6*sin(x)*cos^3(x)] / [cos^2(x) − sin^2(x)]

@loolmath There's already a common denominator, so I'm assuming that you've added the ratios.

[6*sin^3(x)*cos(x) + 6*sin(x)*cos^3(x)] / [cos^2(x) − sin^2(x)]

Now, factor the numerator:

6*sin^3(x)*cos(x) + 6*sin(x)*cos^3(x)

 

[Dr.Peterson]

I'm having a little trouble trying to verify this problem

6 sin^3 (x) cos(x) cos^2(x) − sin^2 (x) + 6 sin(x) cos^3 (x) cos^2(x) − sin^2 (x) = 3 tan(2x)

I kind of understand how to get to 3tan by factoring out the 6 in the top and rewriting the denominator as cos2x but I'm lost on where the 2x in 3tan is coming from

Have you learned the double-angle identities? If not, look them up. They will be very useful here!