Given that sinx=3/5 and x is in quadrant 3, what is the value of tan x/2?

sin (x) = - 3/5

x = arcsin (-3/5)

tan (x/2) = y - - - {Where y is the final answer}

x = 2arctan (y)

Therefore:

arcsin (-3/5) = 2arctan (y)

-3/5 = sin (2arctan (y)) - - - {sin (2x) = 2sin (x) cos (x) }

-3/5 = 2sin (arctan (y)) cos (arctan (y))

-3/10 = sin (arctan (y)) cos (arctan (y))

Construct a triangle on the page. Tan is Opposite / Adjacent, so on the opposite side, you have y, and on the adjacent side, you have 1. By pythagorean theorem, the hypotenuse is sqrt (1 + y²). The sin of this triangle is y / sqrt (1 + y²). The cos is 1 / sqrt (1 + y²)

-3/10 = [y / sqrt (1 + y²) ][1 / sqrt (1 + y²) ]

-3/10 = y / (1 + y²)

(3/10) y² + y + 3/10 = 0

y = - 1/3 or - 3

If you want to check with a calculator, you will know that - 3 is inadmissible. Therefore, - (1/3) is your answer.