Question

Precalculus question I can not find the answer PLEASE HELP AND THANK YOU . right answers only

Answer 1

sin(x)= -3/5, x is in third quadrant

In right angle triangle ,

Opposite side of angle x  = -3

Hypotenuse = 5

Now we find out adjacent side of angle x so that we can find out cosx

We use pythagorean theorem

$hypotenuse^2 = opposite ^2 + adjacent^2$

$5^2 = (-3)^2 + x^2$

25 = 9 +x^2

x^2 = 16, so x= 4

adjacent side = 4

cos (x) =  $\frac{adjacent}{hypotenuse} = \frac{4}{5}$

Cos is negative in third quadrant so cos(x) =  $-\frac{4}{5}$

Now we use double angle formula

$cos(\frac{x}{2})= \sqrt{\frac{1+cosx}{2}} =\sqrt{\frac{1+\frac{4}{5}}{2}}= \frac{\sqrt{90}}{10}$

$sin(\frac{x}{2})= \sqrt{\frac{1-cosx}{2}} =\sqrt{\frac{1-\frac{4}{5}}{2}}= \frac{\sqrt{10}}{10}$

$tan(\frac{x}{2})=\frac{sin(\frac{x}{2})}{cos(\frac{x}{2})}$

$tan(\frac{x}{2})= \frac{\frac{\sqrt{10}}{10}}{\frac{\sqrt{90}}{10} }$

=$\frac{\sqrt{10}}{\sqrt{90}} =\frac{1}{3}$

Tan is positive in third quadrant so final answer is $\frac{1}{3}$