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) =
`(adjacent)/(hypotenuse) = (4)/(5)`

Cos is negative in third quadrant so cos(x) =
`-(4)/(5)`

Now we use double angle formula

`cos((x)/(2))= \sqrt{(1+cosx)/(2)} =\sqrt{(1+(4)/(5))/(2)}= (√(90))/(10)`

`sin((x)/(2))= \sqrt{(1-cosx)/(2)} =\sqrt{(1-(4)/(5))/(2)}= (√(10))/(10)`

`tan((x)/(2))=(sin((x)/(2)))/(cos((x)/(2)))`

`tan((x)/(2))= ((√(10))/(10))/((√(90))/(10) )`

=
`(√(10))/(√(90)) =(1)/(3)`

Tan is positive in third quadrant so final answer is
`(1)/(3)`