Prove the following trig identity In comments to make it clearer
\[\frac{ \sin 3x+\sin x }{ \sin 3x-\sin x }=\frac{ 2 }{ 1-\tan ^{2}x }\]
@jhonyy9 can you help me with this one too?
so depend from what part you wan to go on what part ?
sin3x=sin(2x+x) = ?
can you calculi it using formula sin(a+b) = ?
Yeah I tried that but it became hideously complicated, so I stopped with that Is there a more efficient way that you know of?
I think there is no simpler way, I did it by first apllying sin(a+b) formula, then reducing terms to sines, and at the end dividing by cos^2x to get the right hand side. If you want calculations, just let me know :)
yeap.... I checked the RHS, no dice, it's about just as lengthy
Alright thanks @Fifciol
And yeah @jdoe0001 working the RHS would have too many different processes to unsimplify it and get the answer you need
Hey @fifciol I'm stuck at -4sin^2x+4/-4sin^2+3 can you help?
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