I need to make it equal

[Valentas]

\(\displaystyle sinacos^3a + cosasin^3a = \frac{1}{4}sin4a\)

I was simplyfying left side but I always ended up again on the same left side lol

 

[soroban]

Hello, Valentas!

There must be a typo.

\(\displaystyle \sin\theta \cos^3\!\theta - \cos\theta\sin^3\!\theta \:=\: \tfrac{1}{4}\sin4\theta\)
. . . . . . . . . . . ↑

\(\displaystyle \sin\theta\cos^3\!\theta - \cos\theta\sin^3\!\theta \;=\;\sin\theta\cos\theta(\cos^2\!\theta - \sin^2\!\theta)\)

. . . . . . . .. . . . . . . . . . . . . . \(\displaystyle =\;\left(\tfrac{1}{2}\sin2\theta\right)\left(\cos2\theta) \)

. . . . . . . . . . . . . . . . . . . . . .\(\displaystyle =\;\tfrac{1}{2}\left(\tfrac{1}{2}\sin4\theta\right)\)

. . . . . . . . . . . . . . . . . . . . . .\(\displaystyle =\;\tfrac{1}{4}\sin4\theta\)

 

[Valentas]

Hello, Valentas!

There must be a typo.



\(\displaystyle \sin\theta\cos^3\!\theta - \cos\theta\sin^3\!\theta \;=\;\sin\theta\cos\theta(\cos^2\!\theta - \sin^2\!\theta)\)

. . . . . . . .. . . . . . . . . . . . . . \(\displaystyle =\;\left(\tfrac{1}{2}\sin2\theta\right)\left(\cos2\theta) \)

. . . . . . . . . . . . . . . . . . . . . .\(\displaystyle =\;\tfrac{1}{2}\left(\tfrac{1}{2}\sin4\theta\right)\)

. . . . . . . . . . . . . . . . . . . . . .\(\displaystyle =\;\tfrac{1}{4}\sin4\theta\)

Just solved it like you did. Thank you. I was totally forgot about sin2a

 

[mmm4444bot]

totally forgot about sin(2a)

Hey Valentas, you might have posted your work, instead. That way, we can help guide you with hints, enabling you to finish the exercise yourself versus having us do it for you.

By doing the exercise yourself, perhaps you would not "totally forget about sin(2a)" in the future. :cool:

HMMM, LOOKS LIKE TED BOUGHT ME A PAIR OF BLUE-REFLECTIVE SHADES. NOW, I LOOK EVEN COOLER!