I really need an answer-Derive trig identity

[Dainy]

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(sin2x+sin4x)/(cos2x+cos4x)=tan3x

 

[VietDao29]

You can use these to prove that trig identity:
\cos \alpha + \cos \beta = 2 \cos \frac{\alpha + \beta}{2} \cos \frac{\alpha - \beta}{2}
\cos \alpha - \cos \beta = - 2 \sin \frac{\alpha + \beta}{2} \sin \frac{\alpha - \beta}{2}
\sin \alpha + \sin \beta = 2 \sin \frac{\alpha + \beta}{2} \cos \frac{\alpha - \beta}{2}
\sin \alpha - \sin \beta = 2 \cos \frac{\alpha + \beta}{2} \sin \frac{\alpha - \beta}{2}
Choose 2 of the above equations, and use them.
Can you go from there?
Viet Dao,

 

[thepatientmental]

To derive the trig identity, we can start by using the double angle formula for sine and cosine:

sin2x = 2sinx*cosx
cos2x = cos^2x - sin^2x

Substituting these values into the original equation, we get:

(2sinx*cosx + sin4x) / (cos^2x - sin^2x + cos4x)

Next, we can use the sum and difference identities for sine and cosine to simplify the numerator and denominator:

sin4x = 2sin2x*cos2x
cos4x = cos^2x - sin^2x

Substituting these values, we get:

(2sinx*cosx + 2sin2x*cos2x) / (cos^2x - sin^2x + cos^2x - sin^2x)

Using the double angle formula for cosine, we can further simplify the numerator:

2cosx(2sinx + sin2x) / (cos^2x - sin^2x + 2cos^2x - 2sin^2x)

Now, we can use the Pythagorean identity to replace cos^2x and sin^2x with 1, and simplify the denominator:

2cosx(2sinx + sin2x) / (1 + 2cos^2x - 2sin^2x)

Using the double angle formula for sine, we can simplify the numerator even further:

2cosx(2sinx + 2sinx*cosx) / (1 + 2cos^2x - 2sin^2x)

Finally, we can use the double angle formula for cosine one more time to simplify the denominator and get our desired result of tan3x:

2cosx(2sinx + 2sinx*cosx) / (2cos2x)

Simplifying further, we get:

4sinx(cosx + sinx) / 2cos2x

And finally, we can simplify the fraction by dividing both the numerator and denominator by 2 to get:

2sinx(cosx + sinx) / cos2x = tan3x

Therefore, the derived trig identity is:

(sin2x+sin4x) / (cos2x+cos4x) = tan3x