Given sin x= -4/5 and x is in quadrant 3, what is the value of tan x/2?

Question

I’ve been struggling with this for 20 min any help would be much appreciated! 

Answer 1

In Quad. III both sin and cos are negative. Because tan is sin/cos, it will be positive in Quad. III.

Sine = opp./hyp--->opp. = -4, hyp. = 5 (the hyp. is always positive)

hyp.² = opp.² + adj.²--->adj.² = hyp.² - opp.²--->adj.² = 5² - (-4)²--->adj.² = 25 - 16--->adj. = ±3.

Therefore, sinx=-4/5 and cosx=-3/5.

The half-angle identity for tangent is

                                                     tan(x/2)=(1-cosx)/sinx

Putting in the values we know for sin and cos we get

                                                     tan(x/2)=(1-(-3/5))/(-4/5)

                                                     tan(x/2)=(1+3/5)/(-4/5)

                                                     tan(x/2)=(8/5)/(-4/5)

                                                     tan(x/2)=(8/5)(-5/4)

                                                     tan(x/2)=-2

The answer is -2.

Answer 2

Sin is negative in quadrants 3 and 4.

 

X = inverse_sine(4/5) = 53.130102354156.....

Given the angle is in quadrant 3, it MUST be

 

  180+53.130102354156 = 233.130102354156

 

the tangent of half this angle is then almost -2