Value of Tangent?!?!?

Question

If sin x= 4/5 and x is in quadrant 2, find the value of tan x/2.

Please help! I am not sure of what formula I am supposed to use and how to solve this problem! Thanks :]

10 points

Answer 1

sin x = 4/5

sin = opposite / hypotenuse

so, we know that

opposite = 4

hypotenuse = 5

Using pythagorean theorem, we can find "adjacent"

(opp)^2 + (adj)^2 = (hyp)^2

4^2 + (adj)^2 = 5^2

(adj)^2 = 3

adj = -3 or +3

Since it's quadrant 2, adjacent has to be negative, so adj = -3

so...

cos x = adj / hyp = -3/5

The formula for half angles for tan is

tan (x/2)

= sin x / (1 + cos x)

= (4/5) / (1 + -3/5)

= (4/5) / (2/5)

= 4/5 * 5/2

= 2

Answer 2

what I did is found x...

sin-1 (4/5) = 53.13...

but x is in Q2...

180 - 53.13 = 126.87...

that's x.

now, tan (126.87/2) = 2

I now there is an identity for tan x/2 but I forgot what it is

Answer 3

You will need to use Pythagoreans theorem to solve this. To find tangent, you first need to find all the sides of the triangle. With  sin x= 4/5, you know that the opposite side is 4, and the hypotenuse is 5.

a^2 + b^2 = c^2

Now we need to find the other side, b. Just plug in what we already found and solve:

4^2 + b^2 = 5^2

16 + b^2 = 25

b^2 = 9

b = 3

Now that you know b is equal to 3, we can find tan x/2. Remember, tan x/2  = sin x / 1+cos x.

tan x/2  = sin x / 1+cos x

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

I hope that helps!