(sin 3α/sin α) - (cos 3α/cosα) =2
The Attempt at a Solution
I know for sin 2 α I would put 2 sinαcosα, so for 3α, do I just put 3sinαcosα?
for cos 3α, I'm sorta clueless because there's 3 we can use for cosine,
Then after that step, I know to get both of them on the LHS to have a common denominator, which sinα cosα, please help. Thank yyou in advance!