Ok so basically if tan(α) = m (α =/= 90 degrees) Prove that: a*sin2(α) + b*sin(α)cos(α) + c*cos2(α) = (a*m^2 + b*m + c) / m^2 + 1 a, b, c are parameters That's about it. A pretty simple equation in the sense that this is high school stuff but it could be really tricky if you're not experienced enough with shitty equations like this one (like me). Any help appreciated, thanks in advance!!

You could have shown the other excersices, I wanted to get some fun :[ What is this for? Analityc geometry? Calculus? or high school maths? whatever. asin²α + bsinαcosα + ccos²α cos²α = 1 - sin²α sinαcosα = (1/2)sin(2α) asin²α + (b/2)sin(2α)+c(1-sin²α) (a-c)sin²α + (b/2)sin(2α)+c tgα = m knowing that α =/= 90º (well, generalized, it can't be 90*(2n+1)º ) sinα must be expressed in tgα sinα = tgα/sqrt(1+tg²α) sinα = m/sqrt(1+m²) sin²α = m²/(1+m²) sin(2α) = 2tgα/(1+tg²α) sin(2α) = 2m/(1+m²) (a-c)(m²/(1+m²)+(b/2)*(2m/1+m²)+c (am²-cm²)/(1+m²) + bm/(1+m²)+c (am²-cm²+bm)/(1+m²) + c (am²-cm²+bm+c[1+m²])/(1+m²) (am²-cm²+bm+c+cm²)/(1+m²) (am²+bm+c)/(1+m²) Easy. Any doubt?

Hey, sorry for a kinda late reply First - meh, on the photo you can't see anything else anyway and the most challenging problem on the page is this one so yeah. Secondly, this is high school stuff (as I mentioned in the OP). I have no idea what calculus is (or any college stuff for that matter) Third, thank you!! This took me some time and I didn't do it exactly the way you did it but in the end it all boils down to the basics: Express everything via tangens, and that you do with some formulas such as "sinα = tgα/sqrt(1+tg²α)" (that was unknown to me until some minutes ago, I had to write out proof myself to see how you derived that). So pretty much the secret to solving the problem is the proper usage of the 2 main properties: ( sin2x + cos2x = 1) and (tgx = sinx/cosx). Then it looks like you can go different ways about proving the actual equation, such as using ready formulas like "sinα = tgα/sqrt(1+tg²α)" or just replacing on the spot like I did the first time I solved it (an hour ago or so). Anyway thanks again dude, I understand the problem now and all of these equations and stuff that y'all more experienced people use are a real eye-opener to me as I have to prove them everytime cuz I see them for the first time. Btw I still don't know why "sinαcosα = (1/2)sin(2α)" but you can do the problem without it anyway I'll have to google that one I guess lol Ty for the quick reply too, you da man

sin(2α) = sin(α+α) = sin(α)cos(α)+sin(α)cos(α) = 2sin(α)cos(α) sin(2α) = 2sin(α)cos(α) (1/2)sin(2α) = sin(α)cos(α) It's not that hard, you're welcome ^^