# Hi! I need help to get the sin(x) and tan(x) for this funktion. `cos(x)=-1/sqrt(3)` , `pi/2` < x < `pi` and then count the funktions value for sin(x) and cos(x) from upper...

Hi!

I need help to get the sin(x) and tan(x) for this funktion.

`cos(x)=-1/sqrt(3)` , `pi/2` < x < `pi`

and then count the funktions value for sin(x) and cos(x) from upper funktions answers.

cos (`2x+7pi/6)`

### 4 Answers | Add Yours

You need to evaluate `sin x` using the Pythagorean identity, such that:

`sin x = +-sqrt(1 - cos^2 x)`

Since the problem provides the information that `pi/2 < x < pi,` yields `sin x > 0` , such that:

`sin x = +sqrt(1 - cos^2 x) => sin x = sqrt(1 - (-1/sqrt3)^2)`

`sin x = sqrt(1 - 1/3) => sin x = sqrt(2/3) => sin x = (sqrt6)/3`

You need to use the following trigonometric identity to evaluate `tan x` , such that:

`tan x = sin x/cos x => tan x = ((sqrt2)/(sqrt 3))/(-1/sqrt3)`

Reducing duplicate factors yields:

`tan x = -sqrt2`

You need to evaluate the expression `cos(2x +(7pi)/6)` , using the values of sin x and cos x previously calculated.

You need to expand the cosine of summation, such that:

`cos(2x +(7pi)/6) = cos 2x*cos((7pi)/6) - sin 2x*sin((7pi)/6)`

`cos 2x = cos^2 x - sin^2 x = 1/3 - 2/3 = -1/3`

`sin 2x = 2 sin x*cos x = 2*(sqrt(2/3))*(-1/sqrt3) = -2sqrt2/3`

`cos ((7pi)/6) = cos (pi + pi/6) = -cos(pi/6) = -sqrt3/2`

`sin((7pi)/6) = -sin(pi/6) = -1/2`

`cos(2x +(7pi)/6) = (-1/3)*( -sqrt3/2) + (2sqrt2/3)*(-1/2)`

`cos(2x +(7pi)/6) = (sqrt3 - 2sqrt2)/6`

**Hence, evaluating `sin x` and `tan x` , under the given conditions, yields `sin x = sqrt(2/3)` and `tan x = -sqrt2` and evaluating the expression `cos(2x +(7pi)/6)` yields **`cos(2x +(7pi)/6) = (sqrt3 - 2sqrt2)/6.`

**Sources:**

THANKS ALOT, YOU ARE A KING.

**I mean cos (2x+7pi/6)**

The Second functios is cos (2x + 7`pi`