# 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**

**(sin Θ+sinΘ/2)/1+cosΘ+cosΘ/2 = t**

i've tried splitting so that it is

sin**Θ/1+cos Θ + (sinΘ/2)/(cosΘ/2)**

**which then becomes**

**sin Θ/1+cosΘ + t**

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

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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.**