sinA + tanA = p
sinA + sinA/cosA = p
times cosA
sinAcosA + sinA = pcosA
sinAcosA = pcosA - sinA
but sin 2A = 2sinAcosA
= 2pcosA - 2sinA
??
If sinA+tanA=p then sin2A?
3 answers
if x = sinA
tanA = x/√(1-x^2)
cosA = √(1-x^2)
x + x/√(1-x^2) = p
x^2/(1-x^2) = (p-x)^2
quartics don't solve simply. If you go to wolframalpha . com and type
solve x^2/(1-x^2) = (p-x)^2
you will see a horrendous solution in terms of p.
Anyway, pick a solution and evaluate
2x√(1-x^2)
tanA = x/√(1-x^2)
cosA = √(1-x^2)
x + x/√(1-x^2) = p
x^2/(1-x^2) = (p-x)^2
quartics don't solve simply. If you go to wolframalpha . com and type
solve x^2/(1-x^2) = (p-x)^2
you will see a horrendous solution in terms of p.
Anyway, pick a solution and evaluate
2x√(1-x^2)
tu bta sale
1 answer