If sinA+tanA=p then sin2A?
sinA + tanA = p
sinA + sinA/cosA = p
times cosA
sinAcosA + sinA = pcosA
sinAcosA = pcosA - sinA
but sin 2A = 2sinAcosA
= 2pcosA - 2sinA
??
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)
tu bta sale
To find the value of sin2A, we can use the double-angle formula for sine:
sin2A = 2 * sinA * cosA
Let's find the values of sinA and cosA using the given equation sinA + tanA = p.
Since we know that tanA = sinA / cosA (tangent is the ratio of sine to cosine), we can substitute this into the equation:
sinA + sinA / cosA = p
To simplify the equation, we can multiply through by cosA:
(sinA * cosA) + sinA = p * cosA
Now, let's use the Pythagorean identity sin²A + cos²A = 1 to replace sin²A in the equation:
(1 - cos²A) + sinA = p * cosA
Rearranging the equation:
sinA + (1 - cos²A) = p * cosA
Expanding the equation:
sinA + 1 - cos²A = p * cosA
Rearranging again:
sinA = p * cosA - 1 + cos²A
Now, let's substitute sinA back into the equation for sin2A:
sin2A = 2 * sinA * cosA
sin2A = 2 * (p * cosA - 1 + cos²A) * cosA
simplify:
sin2A = 2p * cos²A - 2cosA + 2cos³A
Therefore, the value of sin2A is 2p * cos²A - 2cosA + 2cos³A.