Proving Sin 6x Cos 4x + Cos 4x sin 2x = Cos 2x tan 8x

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SUMMARY

The equation Sin 6x Cos 4x + Cos 4x sin 2x = Cos 2x tan 8x can be proven using trigonometric identities. The expression Sin 6x Cos 4x simplifies to 1/2(Sin 10x + Sin 2x), while Cos 4x sin 2x simplifies to 1/2(Sin 6x - Sin 2x). Combining these results leads to the expression 1/2(Sin 10x + Sin 6x), which can be further manipulated to show the relationship with Cos 2x tan 8x. The discussion also touches on a personal discovery regarding the addition and subtraction of tangent functions.

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texas_kiwi
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I am trying to prove this equation:

Sin 6x Cos 4x + Cos 4x sin 2x =
Cos 2x tan 8x
Tan 8x​
does anyone have any idea??
 
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[tex]\sin 6x\cos 4x = \frac{1}{2}(\sin 10x + \sin 2x)[/tex]
[tex]\cos 4x\sin 2x = \frac{1}{2}(\sin 6x - \sin 2x)[/tex].

So we get [tex]\frac{1}{2}(\sin 10x + \sin 6x)[/tex]

or [tex]\sin 8x\cos 2x[/tex].

The second expression is not right. Just plug in some values.
 
Last edited:
yes I will tray to slove it

before some month I one like this and I see new law
tan(a)+tan(b)
tab(a)-tan(b)

I discoverd new law juist in my opinion
 

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