If sin theta = 1/5 and theta is in Quadrant I, find the exact value of sin two theta

If sin theta = 1/5 and theta is in Quadrant I, find the exact value of sin two theta

sin 2Ө = 2 sin Ө cos Ө = 2 (1/5) (√24/5)

sin 2Ө = (2/25)(2√6) = 4√6 / 25

Pleased to help----thank you.

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sinx = 1/5
so cosx = ±√(1 - 1/25) -----Pythagorean Identity

cosx = (2/5)√6 > 0 because x in Q1

sin(2x) = 2sinxcosx
sin(2x) = 2*(1/5)*(2/5)√6

sin(2x) = (4/5)√6

sin(t) = 1/5
cos(t) = +/- sqrt(1 - sin(t)^2) = +/- sqrt(24/25) = +/- (2/5) * sqrt(6)

Since we're in Q2, cos(t) = (-2/5) * sqrt(6)


sin(2t) =>
2 * sin(t) * cos(t) =>
2 * (1/5) * (-2/5) * sqrt(6) =>
(-4/25) * sqrt(6)