## Trending News

# can you explain my mistake on u= tan(x/2) on how to proof sin(x)= 2u/1+u^2?

set u=tan(x/2), then arctanU= x/2, then 2arctan u= x

Therefore, sin( 2arctanU0= sin(x), write arctan arctan U in term of Thelta

Draw the triangle, U is the opposite, 1 is adjasent, and Squareroot of U^2 +1 is hypotnews.

From the triangle, we can read sin(theta)= u/sqr of u^2 +1, and cos(theta)= 1/sqr u^2 +1, and then use the formula sin(x)= 2sin(theta)cos(theta)=2 cos(theta)sin(theta)= sin(x) which is 2u/1+u^2

Reply to the user below: according to my professor, i cant assume that arctan U= x/2, because if x=400 pi, then u= tan( 400pi/2)= tan(200pi)=o, but arctan(0) is not equal to 400pi/2!!!!!

### 1 Answer

- 7 years agoFavorite Answer
u = tan(x/2)

1/u = cot(x/2)

(1/u)^2 = cot(x/2)^2

(1/u)^2 = csc(x/2)^2 - 1

1 + (1/u)^2 = csc(x/2)^2

(1 + u^2) / u^2 = csc(x/2)^2

u^2 / (1 + u^2) = sin(x/2)^2

u^2 / (1 + u^2) = (1/2) * (1 - cos(x))

2 * u^2 / (1 + u^2) = 1 - cos(x)

cos(x) = 1 - 2u^2 / (1 + u^2)

cos(x) = (1 + u^2 - 2u^2) / (1 + u^2)

cos(x) = (1 - u^2) / (1 + u^2)

sqrt(1 - sin(x)^2) = (1 - u^2) / (1 + u^2)

1 - sin(x)^2 = (1 - u^2)^2 / (1 + u^2)^2

1 - (1 - 2u^2 + u^4) / (1 + 2u^2 + u^4) = sin(x)^2

(1 + 2u^2 + u^4 - 1 + 2u^2 - u^4) / (1 + u^2)^2 = sin(x)^2

sin(x)^2 = 4u^2 / (1 + u^2)^2

sin(x) = 2u / (1 + u^2)

If I'm reading your steps correctly

u = tan(x/2)

arctan(u) = x/2

2 * arctan(u) = x

sin(2 * arctan(u)) = sin(x)

sin(x) = 2 * sin(arctan(u)) * cos(arctan(u))

sin(arctan(u)) = u / sqrt(1 + u^2)

cos(arctan(u)) = 1 / sqrt(1 + u^2)

sin(x) = 2 * u * 1 / (1 + u^2)

sin(x) = 2u / (1 + u^2)

Looks like you did it right to me. Why are you saying there's a mistake?