# If tan Θ/2 = t, express each of the following in terms of t (sinΘ+sinΘ/2)/1+cosΘ+cosΘ/2 = t   i've tried splitting so that it is sinΘ/1+cosΘ + (sinΘ/2)/(cosΘ/2) which then becomes...

If tan Θ/2 = t, express each of the following in terms of t

(sinΘ+sinΘ/2)/1+cosΘ+cosΘ/2 = t

i've tried splitting so that it is

sinΘ/1+cosΘ + (sinΘ/2)/(cosΘ/2)

which then becomes

sinΘ/1+cosΘ + t

im pretty sure sinΘ/1+cosΘ is equal to zero, but not sure what to do

sciencesolve | Certified Educator

You need to write `sin theta`  using the formula of double angle such that:

`sin theta = sin 2(theta/2)`  (notice that if you reduce by 2 the value of `sin theta`  is preserved)

Hence, `sin theta = 2 sin (theta/2)*cos (theta/2)`

Use this formula at numerator such that:

`sin theta + sin (theta/2) = 2 sin (theta/2)*cos (theta/2) + sin (theta/2)`

Factoring out `sin (theta/2)`  yields:

`sin (theta/2)*(2 cos(theta/2) + 1)`

You need to write cos theta using the formula of double angle such that:

`cos theta = 2 cos^2 (theta/2) - 1`

Writing the new form of denominator yields:

`1 + cos theta + cos(theta/2) = 1 + 2 cos^2 (theta/2) - 1 + cos(theta/2)`

`` `1 + cos theta + cos(theta/2) = 2 cos^2 (theta/2) + cos(theta/2)`

Factoring out `cos(theta/2)`  yields:

`cos(theta/2)(2 cos (theta/2) + 1)`

You need to write the new form of expression such that:

`(sin (theta/2)*(2 cos(theta/2) + 1))/(cos(theta/2)(2 cos (theta/2) + 1)) = t`

`tan(theta/2)*(2 cos(theta/2) + 1)/(2 cos (theta/2) + 1) = t`

Plugging `tan (theta/2) = t`  in expression yields:

`t*(2 cos(theta/2) + 1)/(2 cos (theta/2) + 1) = t`

`` `(2 cos(theta/2) + 1)/(2 cos (theta/2) + 1) = 1`

1= 1

Hence, using `tan(theta/2)=t`  yields 1=1, hence the identity `(sin theta + sin(theta/2))/(1 + cos theta + cos(theta/2)) = t`  is established.