Cot=3 Find sin(x/2), cos(x/2), tan(x/2)

Find sin(x/2), cos(x/2), and tan(x/2) from the given information.
cot(x) = 3, 180 < x < 270

Please explain the steps you take!

cot x = 3
tan x = 1/3
since x is in the third quadrant,
sin x = -1/√10, cos x = -3/√10

x/2 is in the second quadrant...

tan(x/2) = sin x / (1 + cos x) = -1/(√10 - 3) = -3 - √10

sec^2(x/2) = 1 + tan^2(x/2) = 1 + 19 + 6√10 = 20 + 6√10

tan(x/2) = -3 - √10
cos(x/2) = -1/√(20 + 6√10)
sin(x/2) = (3 + √10) / √(20 + 6√10)

Cot(x) = 3 implies
Cos(x)/Sin(x) = 3

180
With the above angles, cos(x) is negative and sin(x) is also negative. So sin(x)*3 = cos(x)

but Cos(x)^2+Sin(x)^2 = 1

so x^2 + y^2 = 1, with x/y = 3.

put a=x y=b, taking a as cos(x) and b as sin(x) so a=3*b

(3*b)^2 + b^2 = 1, 10*b^2 = 1 b^2 = 1/10 b = -1*sqrt(1/10) so x = arcsin(-1*sqrt(1/10))

Sin(x/2) = Sin(arcsin(-1/2*sqrt(1/10)) = -1/2*sqrt(1/10)

cos(x/2) = -sqrt(1 - sin(x/2)^2)= -sqrt(1-1/4*sqrt(1/10)).

tan(x/2) = sin(x/2)/cos(x/2) = -1/2*sqrt(1/10)/(-1*sqrt(1 - 1/4*sqrt(1/10)))

tan(x) = ⅓
2tan(x/2) / [ 1−tan²(x/2) ] = ⅓
6tan(x/2) = 1 − tan²(x/2)
tan²(x/2) + 6tan(x/2) − 1 = 0
tan(x/2) = −3±√10 (use quadratic formula)
180 < x < 270
90 < (x/2) < 135 --- 2nd quadrant where tan(x/2) is negative
tan(x/2) = −3−√10
Hypotenuse = √(20+6√10) (use Pythagoras' Theorem)
Opposite = 3+√10
Adjacent = −1
Hence
sin(x/2) = (3+√10) / √(20+6√10)
cos(x/2) = −1 / √(20+6√10) and
tan(x/2) = −(3+√10).

cot (x) = 1/ tan x

1/tanx = 3

1 = 3 tan x
tan x = 1/3
basic angle = tan^-1(1/3)
basic angle = 18.435

tangent positive is in the1st quadrant and 3rd quadrant.

In the 1st quadrant -> 18.4, this is rejected value because it is out of the given range.

In the 3rd quadrant -> 180 + 18.4 = 198.4, this value is accepted because it is in range.

So, x = 198.4 degree

Substitute x into...

Sin (x/2) = ? , Cos (x/2) = ? . Tan (x/2) = ?

Sin (198.4/2) = 0.99 (2 decimal place)

Cos (198.4/2) = -0.16 (2 decimal place)

Tan (198.4/2) = -6.17 (2 decimal place)