prove trig identity cos(2x) / [3 - sin(2x) - 4sin^2(x)] = [tan(x) + 1] / [tan(x) + 3]

[MrClibert]

Please help me with this problem

cos(2x) / [3 - sin(2x) - 4sin^2(x)] = [tan(x) + 1] / [tan(x) + 3]

 

[topsquark]

Please help me with this problem

Start with the RHS:
Hint 1: Rewrite tan(x) as sin(x)/cos(x) and multiply the numerator and denominator by cos(x).

Hint2: Multiply the numerator and denominator by cos(x) - sin(x).

See what you can do.

Note:. This isn't quite a true identity as it is not defined on the LHS for certain values of x that exist on the RHS. But for all x such that both sides are defined, it is an identity.

-Dan

 

[MarkFL]

Try using double angle identities, then divide numerator and denominator by the square of the cosine function , apply a Pythagorean identity in the denominator to get rid of the secant function, then factor and cancel.

 

[Steven G]

I think that this is the 1st time I have ever disagreed with topsquark. I would think changing the left side would be easier. The lhs has 2x as some of its angle and it would be easier, imo, to convert does terms to having just x as its angles. Give that a try.

 

[topsquark]

I think that this is the 1st time I have ever disagreed with topsquark. I would think changing the left side would be easier. The lhs has 2x as some of its angle and it would be easier, imo, to convert does terms to having just x as its angles. Give that a try.

I did it that way so we wouldn't have to cancel the cos(x) - sin(x) factor. That's the part that "ruins" the exact definition of an identity, so I thought it would be "prettier" to avoid it.

-Dan

 

[MrClibert]

Thank you, everyone.

 

[MarkFL]

This is what I had in mind. Begin with the lhs:

\(\displaystyle \frac{\cos(2x)}{3-\sin(2x)-4\sin^2(x)}\)

Apply double-angle identities:

\(\displaystyle \frac{\cos^2(x)-\sin^2(x)}{3-2\sin(x)\cos(x)-4\sin^2(x)}\)

Divide numerator and denominator by \(\displaystyle \cos^2(x)\)

\(\displaystyle \frac{1-\tan^2(x)}{3\sec^2(x)-2\tan(x)-4\tan^2(x)}\)

In the denominator, apply a Pythagorean identity:

\(\displaystyle \frac{1-\tan^2(x)}{3\tan^2(x)+3-2\tan(x)-4\tan^2(x)}\)

\(\displaystyle \frac{\tan^2(x)-1}{\tan^2(x)+2\tan(x)-3}\)

\(\displaystyle \frac{(\tan(x)+1)(\tan(x)-1)}{(\tan(x)+3)(\tan(x)-1)}\)

Divide out common factor:

\(\displaystyle \frac{\tan(x)+1}{\tan(x)+3}\)

 

[MrClibert]

Thank you for your responses.

 

[MrClibert]

This is what I had in mind. Begin with the lhs:

\(\displaystyle \frac{\cos(2x)}{3-\sin(2x)-4\sin^2(x)}\)

Apply double-angle identities:

\(\displaystyle \frac{\cos^2(x)-\sin^2(x)}{3-2\sin(x)\cos(x)-4\sin^2(x)}\)

Divide numerator and denominator by \(\displaystyle \cos^2(x)\)

\(\displaystyle \frac{1-\tan^2(x)}{3\sec^2(x)-2\tan(x)-4\tan^2(x)}\)

In the denominator, apply a Pythagorean identity:

\(\displaystyle \frac{1-\tan^2(x)}{3\tan^2(x)+3-2\tan(x)-4\tan^2(x)}\)

\(\displaystyle \frac{\tan^2(x)-1}{\tan^2(x)+2\tan(x)-3}\)

\(\displaystyle \frac{(\tan(x)+1)(\tan(x)-1)}{(\tan(x)+3)(\tan(x)-1)}\)

Divide out common factor:

\(\displaystyle \frac{\tan(x)+1}{\tan(x)+3}\)

Nice! Thank you.